Quadratic Equations for JEE: Roots, Graphs, Location and Newton Sums

A quadratic equation is the first equation where the answer is not a single number but a pair, and the pair carries more information than either root alone. Their sum and product come straight from the coefficients, their nature comes from one number $D$, and their position on the number line comes from one picture, the parabola. JEE uses all three in almost every paper, and the same ideas return in calculus, coordinate geometry and complex numbers.

This note covers the equation and its roots, the discriminant, the relations between roots and coefficients, forming and transforming equations, power sums by Newton's formula, the graph and sign of the quadratic expression, maximum, minimum and range, location of roots, common roots, equations reducible to a quadratic, the quadratic expression in two variables, and equations of higher degree. The worked examples include JEE Main 2025 and 2026 questions and JEE Advanced questions; every answer was re-derived from scratch.

The quadratic equation and its roots

An equation of the form

$$ax^2+bx+c=0,\qquad a\ne0$$

with $a,b,c$ real is a quadratic equation in $x$. A number $\alpha$ is a root (or zero) when $a\alpha^2+b\alpha+c=0$. A quadratic equation has exactly two roots, counted with repetition; they may be real or non-real.

Degenerate cases. If $a=0$, $b\ne0$, the equation is linear with the single root $-\tfrac cb$. If $a=b=0$, $c\ne0$, there is no root. If $a=b=c=0$, every $x$ satisfies it.

For example, for distinct $a,b,c$ the expression

$$\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}$$ $$+\frac{(x-c)(x-a)}{(b-c)(b-a)}-1$$

has degree at most $2$ and is $0$ at $x=a$, $x=b$ and $x=c$ (check each). So it is $0$ for every $x$.

Methods of solving

Factorisation. Split the middle term so that $ax^2+bx+c=a(x-\alpha)(x-\beta)$, then put each factor equal to zero.

Completing the square. Divide by $a$ and add $\left(\tfrac{b}{2a}\right)^2$ to both sides:

$$x^2+\frac bax=-\frac ca$$ $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$

Taking square roots gives the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The number $D=b^2-4ac$ is the discriminant. The two roots are

$$\alpha=\frac{-b+\sqrt D}{2a},\qquad \beta=\frac{-b-\sqrt D}{2a}$$

Example 1 (IITianForum). Solve (a) $6x^2-7x-3=0$ by factorisation; (b) $2x^2+4x-1=0$ by the formula.

▸ SHOW SOLUTION

Step 1. Part (a)

Two numbers with product $6\cdot(-3)=-18$ and sum $-7$ are $-9$ and $2$:

$$6x^2-9x+2x-3=(2x-3)(3x+1)$$

So $x=\tfrac32$ or $x=-\tfrac13$.

Step 2. Part (b)

$D=16+8=24$:

$$x=\frac{-4\pm\sqrt{24}}{4}=\frac{-2\pm\sqrt6}{2}$$

Answer: (a) $\tfrac32,\ -\tfrac13$; (b) $\dfrac{-2\pm\sqrt6}{2}$

Example 2 (JEE Main 2026, 23 January Shift 1). A building construction work can be completed by two masons $A$ and $B$ together in $22.5$ days. Mason $A$ alone can complete the work in $24$ days less than mason $B$ alone. Then mason $A$ alone will complete the work in

▸ SHOW SOLUTION

Step 1. Form the equation

Let $A$ take $x$ days; then $B$ takes $x+24$ days. One day's work together:

$$\frac1x+\frac1{x+24}=\frac{2}{45}$$

Step 2. Solve

$$45(2x+24)=2x(x+24)$$$$x^2-21x-540=0$$$$(x-36)(x+15)=0$$

A number of days is positive, so $x=36$.

Answer: option (4)

Example 3 (JEE Main 2026, 21 January Shift 2). The positive integer $n$, for which the solutions of the equation

$$x(x+2)+(x+2)(x+4)+\cdots$$ $$\cdots+(x+2n-2)(x+2n)=\frac{8n}3$$

are two consecutive even integers, is

▸ SHOW SOLUTION

Step 1. Add the terms

The $r$th term is $x^2+(4r-2)x+4r^2-4r$. Summing for $r=1$ to $n$ and dividing by $n$:

$$x^2+2nx+\frac{4(n^2-1)}{3}-\frac83=0$$

Step 2. Roots differ by 2

For a monic quadratic, $(\alpha-\beta)^2=D$, so $D=4$:

$$4n^2-\frac{16(n^2-1)}{3}+\frac{32}{3}=4$$$$n^2=9,\qquad n=3$$

Step 3. Check

$n=3$ gives $x^2+6x+8=0$, roots $-2,-4$: consecutive even integers.

Answer: option (1)

Example 4 (JEE Main 2026, 2 April Shift 1). Let $\alpha$ and $\alpha+2$, $\alpha\in\mathbb Z$, be the roots of

$$x(x+2)+(x+1)(x+3)+\cdots$$ $$\cdots+(x+n-1)(x+n+1)=4n$$

for some $n\in\mathbb N$. Then $n+\alpha$ is equal to

▸ SHOW SOLUTION

Step 1. Add the terms

The $k$th term is $(x+k)^2-1=x^2+2kx+k^2-1$. Summing and dividing by $n$:

$$x^2+(n+1)x+C=0$$$$C=\frac{(n+1)(2n+1)}6-5$$

Step 2. Roots differ by 2

$D=(n+1)^2-4C=4$:

$$(n+1)^2-\frac{2(n+1)(2n+1)}{3}+20=4$$$$-n^2+61=12,\qquad n=7$$

Step 3. The roots

$n=7$ gives $x^2+8x+15=0$, roots $-5,-3$. So $\alpha=-5$.

Answer: $n+\alpha=2$, option (3)

Nature of the roots

With $a,b,c$ real, $\sqrt D$ in the formula decides everything.

With $a,b,c$ rational:

Why irrational roots come in pairs. Let $p+\sqrt q$ be a root ($\sqrt q$ irrational). The other root is $-\tfrac ba-(p+\sqrt q)$, which is $p'-\sqrt q$ with $p'$ rational. The product $(p+\sqrt q)(p'-\sqrt q)=\tfrac ca$ is rational, so $(p'-p)\sqrt q$ is rational, which forces $p'=p$.

Integer roots. If $a=1$ and $b,c$ are integers, the roots are integers exactly when $D$ is a perfect square. (Here $D=b^2-4c$ has the same parity as $b^2$, so $-b\pm\sqrt D$ is even.)

Example 5 (IITianForum). Find all real $k$ for which $x^2-2(k+1)x+k^2+3=0$ has real roots. When are they equal?

▸ SHOW SOLUTION

Step 1. Discriminant

$$\frac D4=(k+1)^2-(k^2+3)=2k-2$$

Step 2. Conditions

Real roots: $2k-2\ge0$, so $k\ge1$. Equal roots: $k=1$, and then the root is $k+1=2$.

Answer: $k\ge1$; equal roots at $k=1$

Example 6 (JEE Main 2025, 29 January Shift 2). If the set of all $a\in\mathbb R$ for which the equation $2x^2+(a-5)x+15=3a$ has no real root is the interval $(\alpha,\beta)$, and $X=\{x\in\mathbb Z:\alpha<x<\beta\}$, then $\sum_{x\in X}x^2$ is equal to

▸ SHOW SOLUTION

Step 1. No real root

$$(a-5)^2-8(15-3a)<0$$$$a^2+14a-95<0$$$$(a+19)(a-5)<0$$

So $(\alpha,\beta)=(-19,5)$.

Step 2. The sum

$X=\{-18,\dots,4\}$:

$$\sum x^2=\frac{18\cdot19\cdot37}{6}+30=2139$$

Answer: option (3)

Example 7 (JEE Main 2025, 23 January Shift 1). If the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has equal roots, where $a+c=15$ and $b=\tfrac{36}5$, then $a^2+c^2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. One root is 1

The coefficients add to $0$:

$$a(b-c)+b(c-a)+c(a-b)=0$$

So $x=1$ is a root, and equal roots means both roots are $1$.

Step 2. Product of roots is 1

$$c(a-b)=a(b-c)\ \Rightarrow\ 2ac=b(a+c)$$$$2ac=\frac{36}{5}\cdot15=108,\qquad ac=54$$

Step 3. Value

$$a^2+c^2=225-108=117$$

Answer: $117$

Example 8 (JEE Main 2025, 4 April Shift 1). Consider $x^2+4x-n=0$, where $n\in[20,100]$ is a natural number. The number of values of $n$ for which the equation has integral roots is

▸ SHOW SOLUTION

Step 1. Complete the square

$$(x+2)^2=n+4,\qquad x=-2\pm\sqrt{n+4}$$

Integral roots need $n+4$ to be a perfect square.

Step 2. Count

$24\le n+4\le104$ allows $25,36,49,64,81,100$: six values.

Answer: option (3)

Example 9 (JEE Advanced 2026, Paper 2). Let $a,b,c$ be positive integers in arithmetic progression such that $ax^2+bx+c=0$ has only integer solutions. Which of the following statements is (are) TRUE?

▸ SHOW SOLUTION

Step 1. Reduce to a monic equation

Integer roots give integer sum $-\tfrac ba$ and product $\tfrac ca$. Put $b=ma$, $c=na$. The A.P. gives $2b=a+c$, so $n=2m-1$:

$$x^2+mx+(2m-1)=0$$

Step 2. Perfect square discriminant

$$D=m^2-8m+4=(m-4)^2-12$$

$D=d^2$ gives $(m-4-d)(m-4+d)=12$. Both factors have the same parity, so they are $2,6$ or $-6,-2$. With $m\ge1$: $m=8$, $d=2$.

Step 3. The equation

$b=8a$, $c=15a$, and $x^2+8x+15=0$ has roots $-3,-5$.

Step 4. Statements

(1)$c-b=7a$: true.

(2)$-3,-5$ are odd: true.

(3)$c=15$ gives $a=1$, $b=8$, $ab=8$: true.

(4)$3$ is not a root: false.

Answer: (1), (2), (3)

Roots and coefficients

If $\alpha,\beta$ are the roots of $ax^2+bx+c=0$, then $ax^2+bx+c=a(x-\alpha)(x-\beta)$. Comparing coefficients:

$$\alpha+\beta=-\frac ba,\qquad \alpha\beta=\frac ca$$

From the formula, the difference is

$$(\alpha-\beta)^2=\frac{D}{a^2},\qquad |\alpha-\beta|=\frac{\sqrt D}{|a|}$$

Write $S=\alpha+\beta$ and $P=\alpha\beta$. Every symmetric expression in the roots (unchanged when $\alpha,\beta$ swap) is a formula in $S$ and $P$:

Signs of the roots (real coefficients):

Example 10 (IITianForum). If $\alpha,\beta$ are the roots of $2x^2-5x+1=0$, find $\alpha^2+\beta^2$, $\alpha^3+\beta^3$, $\tfrac1{\alpha^2}+\tfrac1{\beta^2}$ and $(\alpha-\beta)^2$.

▸ SHOW SOLUTION

Step 1. S and P

$S=\tfrac52$, $P=\tfrac12$.

Step 2. Each expression

$$\alpha^2+\beta^2=\frac{25}4-1=\frac{21}4$$$$\alpha^3+\beta^3=\frac{125}8-\frac{15}4=\frac{95}8$$$$\frac1{\alpha^2}+\frac1{\beta^2}=\frac{21/4}{1/4}=21$$$$(\alpha-\beta)^2=\frac{25}4-2=\frac{17}4$$

Answer: $\tfrac{21}4,\ \tfrac{95}8,\ 21,\ \tfrac{17}4$

Example 11 (JEE Main 2026, 28 January Shift 1). If $\alpha,\beta$ $(\alpha<\beta)$ are the roots of $\lambda x^2-(\lambda+3)x+3=0$ such that $\tfrac1\alpha-\tfrac1\beta=\tfrac13$, then the sum of all possible values of $\lambda$ is

▸ SHOW SOLUTION

Step 1. Use the product

$$\frac1\alpha-\frac1\beta=\frac{\beta-\alpha}{\alpha\beta}=\frac13$$

$\alpha\beta=\tfrac3\lambda$, so $\beta-\alpha=\tfrac1\lambda$.

Step 2. Square and use S, P

$$\frac1{\lambda^2}=\frac{(\lambda+3)^2}{\lambda^2}-\frac{12}{\lambda}$$$$\lambda^2-6\lambda+8=0,\qquad \lambda=2,4$$

Both work: $\lambda=2$ gives roots $1,\tfrac32$; $\lambda=4$ gives $\tfrac34,1$.

Answer: $2+4=6$, option (1)

Example 12 (JEE Main 2026, 22 January Shift 2). Let $\alpha,\beta$ be the roots of $12x^2-20x+3\lambda=0$, $\lambda\in\mathbb Z$. If $\tfrac12\le|\beta-\alpha|\le\tfrac32$, then the sum of all possible values of $\lambda$ is

▸ SHOW SOLUTION

Step 1. The difference

$S=\tfrac53$, $P=\tfrac\lambda4$:

$$(\beta-\alpha)^2=\frac{25}9-\lambda$$

Step 2. The bound

$$\frac14\le\frac{25}9-\lambda\le\frac94$$$$\frac{19}{36}\le\lambda\le\frac{91}{36}$$

So $\lambda=1,2$.

Answer: $3$, option (3)

Example 13 (JEE Advanced 2015). Let $S$ be the set of all non-zero real numbers $\alpha$ such that $\alpha x^2-x+\alpha=0$ has two distinct real roots $x_1,x_2$ with $|x_1-x_2|<1$. Which of these intervals is (are) a subset of $S$?

▸ SHOW SOLUTION

Step 1. Distinct real roots

$D=1-4\alpha^2>0$, so $\alpha^2<\tfrac14$.

Step 2. Difference of roots

$$|x_1-x_2|^2=\frac{1-4\alpha^2}{\alpha^2}<1$$$$1<5\alpha^2,\qquad \alpha^2>\frac15$$

Step 3. Combine

$$S=\left(-\tfrac12,-\tfrac1{\sqrt5}\right)\cup\left(\tfrac1{\sqrt5},\tfrac12\right)$$

Answer: (A) and (D)

Example 14 (JEE Main 2026, 2 April Shift 2). Let $\alpha,\beta$ be the roots of $x^2-3x+r=0$, and $\tfrac\alpha2,\ 2\beta$ be the roots of $x^2+3x+r=0$. If the roots of $x^2+6x=m$ are $2\alpha+\beta+2r$ and $\alpha-2\beta-\tfrac r2$, then $m$ is

▸ SHOW SOLUTION

Step 1. Two sums

$$\alpha+\beta=3,\qquad \frac\alpha2+2\beta=-3$$

So $\beta=-3$, $\alpha=6$, and $r=\alpha\beta=-18$.

Step 2. The new roots

$$2\alpha+\beta+2r=12-3-36=-27$$$$\alpha-2\beta-\frac r2=6+6+9=21$$

Step 3. Product of roots of x² + 6x − m = 0

$$-m=(-27)(21)=-567$$

Answer: $m=567$, option (D)

Example 15 (JEE Main 2026, 5 April Shift 2). Let $\alpha,\beta$ be the roots of $x^2-x+p=0$ and $\gamma,\delta$ the roots of $x^2-4x+q=0$, $p,q\in\mathbb Z$. If $\alpha,\beta,\gamma,\delta$ are in G.P., then $|p+q|$ equals

▸ SHOW SOLUTION

Step 1. Write the G.P.

Take $\alpha=a$, $\beta=ar$, $\gamma=ar^2$, $\delta=ar^3$:

$$a(1+r)=1,\qquad ar^2(1+r)=4$$

Dividing, $r^2=4$.

Step 2. Integer p, q

$r=2$ gives $a=\tfrac13$ and $p=a^2r=\tfrac29$: not an integer. $r=-2$ gives $a=-1$:

$$p=a^2r=-2,\qquad q=a^2r^5=-32$$

Answer: $|p+q|=34$, option (3)

Example 16 (JEE Main 2025, 22 January Shift 2). Let $\alpha_\theta,\beta_\theta$ be the distinct roots of $2x^2+(\cos\theta)x-1=0$, $\theta\in(0,2\pi)$. If $m$ and $M$ are the minimum and maximum values of $\alpha_\theta^4+\beta_\theta^4$, then $16(M+m)$ equals

▸ SHOW SOLUTION

Step 1. In S and P

$S=-\tfrac{\cos\theta}2$, $P=-\tfrac12$. With $c=\cos^2\theta$:

$$\alpha^2+\beta^2=\frac c4+1$$$$\alpha^4+\beta^4=\left(\frac c4+1\right)^2-\frac12$$

Step 2. Extremes

$c\in[0,1]$, and the expression increases with $c$:

$$m=\frac12,\qquad M=\frac{25}{16}-\frac12=\frac{17}{16}$$

Answer: $16(M+m)=25$, option (2)

Example 17 (JEE Main 2026, 6 April Shift 2). For $(n^2-2n+2)x^2-3x+(n^2-2n+2)^2=0$, $n\in\mathbb R$, let $\alpha$ be the minimum value of the product of the roots and $\beta$ the maximum value of the sum of the roots. The sum of the first six terms of the G.P. with first term $\alpha$ and common ratio $\tfrac\alpha\beta$ is

▸ SHOW SOLUTION

Step 1. Sum and product

Let $k=n^2-2n+2=(n-1)^2+1\ge1$. Product $=\tfrac{k^2}{k}=k$, least value $\alpha=1$. Sum $=\tfrac3k$, greatest value $\beta=3$.

Step 2. The G.P.

$$S_6=\frac{1-(1/3)^6}{1-1/3}=\frac{364}{243}$$

Answer: option (3)

Example 18 (JEE Advanced 2016). Let $-\tfrac\pi6<\theta<-\tfrac\pi{12}$. Suppose $\alpha_1,\beta_1$ are the roots of $x^2-2x\sec\theta+1=0$ and $\alpha_2,\beta_2$ are the roots of $x^2+2x\tan\theta-1=0$. If $\alpha_1>\beta_1$ and $\alpha_2>\beta_2$, then $\alpha_1+\beta_2$ equals

▸ SHOW SOLUTION

Step 1. First equation

$$x=\sec\theta\pm\sqrt{\sec^2\theta-1}=\sec\theta\pm\tan\theta$$

Here $\tan\theta<0$, so the larger root is $\alpha_1=\sec\theta-\tan\theta$.

Step 2. Second equation

$$x=-\tan\theta\pm\sec\theta$$

$\sec\theta>0$, so the smaller root is $\beta_2=-\tan\theta-\sec\theta$.

Step 3. Add

$$\alpha_1+\beta_2=-2\tan\theta$$

Answer: option (C)

Example 19 (IITianForum). Let $\alpha,\beta$ be the roots of $x^2+ax-\tfrac1{a^2}=0$, $a$ a non-zero real. Find the least value of $[\alpha^4+\beta^4]$, where $[\cdot]$ is the greatest integer function.

▸ SHOW SOLUTION

Step 1. In S and P

$S=-a$, $P=-\tfrac1{a^2}$:

$$\alpha^2+\beta^2=a^2+\frac2{a^2}$$$$\alpha^4+\beta^4=\left(a^2+\frac2{a^2}\right)^2-\frac2{a^4}$$$$=a^4+4+\frac2{a^4}$$

Step 2. AM ≥ GM

$$a^4+\frac2{a^4}\ge2\sqrt2$$

with equality at $a^4=\sqrt2$. So the least value of $\alpha^4+\beta^4$ is $4+2\sqrt2=6.82\ldots$

Answer: $[4+2\sqrt2]=6$

Forming and transforming equations

The quadratic with roots $\alpha$ and $\beta$ is

$$x^2-(\alpha+\beta)x+\alpha\beta=0$$

that is, $x^2-(\text{sum})x+(\text{product})=0$. Any non-zero multiple has the same roots.

Transforming the roots. Let $\alpha,\beta$ be the roots of $ax^2+bx+c=0$. To get the equation whose roots are $f(\alpha),f(\beta)$, put $y=f(x)$, solve for $x$, substitute in the given equation and rename $y$ as $x$.

Example 20 (IITianForum). If $\alpha,\beta$ are the roots of $2x^2+3x-4=0$, form the equations whose roots are (a) $\tfrac1\alpha,\tfrac1\beta$; (b) $\alpha+2,\beta+2$; (c) $\alpha^2,\beta^2$.

▸ SHOW SOLUTION

Step 1. Reciprocals

Reverse the coefficients: $-4x^2+3x+2=0$, that is $4x^2-3x-2=0$.

Step 2. Shift by 2

$$2(x-2)^2+3(x-2)-4=0$$$$2x^2-5x-2=0$$

Step 3. Squares

$$\alpha^2+\beta^2=\frac94+4=\frac{25}4,\qquad \alpha^2\beta^2=4$$$$4x^2-25x+16=0$$

Answer: $4x^2-3x-2=0$; $2x^2-5x-2=0$; $4x^2-25x+16=0$

Example 21 (IIT JEE 2010). Let $p,q$ be real with $p\ne0$, $p^3\ne q$, $p^3\ne-q$. If $\alpha,\beta$ are non-zero complex numbers with $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$, find a quadratic equation whose roots are $\tfrac\alpha\beta$ and $\tfrac\beta\alpha$.

▸ SHOW SOLUTION

Step 1. Product αβ

$$\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$$$$q=-p^3+3p\,\alpha\beta,\qquad \alpha\beta=\frac{p^3+q}{3p}$$

Step 2. Sum and product of the new roots

The product is $1$. The sum is

$$\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{p^2}{\alpha\beta}-2=\frac{p^3-2q}{p^3+q}$$

Step 3. The equation

$$(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$$

Answer: as above

Example 22 (IITianForum). Let $\alpha,\beta$ be the roots of $x^2-2x-1=0$. Find the quadratic equation whose roots are $\alpha^3-\dfrac1{\alpha^3}$ and $\beta^2+\dfrac1{\beta^2}$.

▸ SHOW SOLUTION

Step 1. A value shared by both roots

Each root satisfies $x^2-2x-1=0$. Divide by $x$:

$$x-\frac1x=2$$

This holds for $x=\alpha$ and for $x=\beta$.

Step 2. The two numbers

$$\alpha^3-\frac1{\alpha^3}=\left(\alpha-\frac1\alpha\right)^3+3\left(\alpha-\frac1\alpha\right)$$$$=8+6=14$$$$\beta^2+\frac1{\beta^2}=\left(\beta-\frac1\beta\right)^2+2=6$$

Step 3. The equation

Sum $20$, product $84$.

Answer: $x^2-20x+84=0$

Example 23 (IITianForum). The equation $(x-a)(x-b)+2(x-c)(x-d)=0$ has roots $p$ and $q$. Find the roots of $3(x-p)(x-q)-(x-a)(x-b)=0$.

▸ SHOW SOLUTION

Step 1. Same roots, same leading coefficient

$(x-a)(x-b)+2(x-c)(x-d)$ has leading coefficient $3$ and roots $p,q$. So for every $x$,

$$(x-a)(x-b)+2(x-c)(x-d)$$$$=3(x-p)(x-q)$$

Step 2. Rearrange

$$3(x-p)(x-q)-(x-a)(x-b)$$$$=2(x-c)(x-d)$$

Answer: $c$ and $d$

Example 24 (JEE Advanced 2026, Paper 1). Match List I with List II.

(P) If $\alpha,\beta$ are distinct roots of $x^2+x+1=0$, the quadratic with roots $\dfrac1{(\alpha+1)^{2026}},\dfrac1{(\beta+1)^{2026}}$ is (Q) If $\alpha,\beta$ are distinct roots of $x^2+x+1=0$, the quadratic with roots $\dfrac1{(\alpha+1)^{2027}},\dfrac1{(\beta+1)^{2027}}$ is (R) If $\gamma,\delta$ are distinct roots of $x^2-x+1=0$, then $\dfrac1{(\gamma-1)^{2026}}+\dfrac1{(\delta-1)^{2026}}$ is (S) If $p,r$ are distinct roots of $x^2+x-1=0$, then $\dfrac1{(p+1)^3}+\dfrac1{(r+1)^3}$ is

List II: (1) $x^2+x+1=0$ (2) $x^2-x+1=0$ (3) $x^2+x-1=0$ (4) $-1$ (5) $-4$

▸ SHOW SOLUTION

Step 1. Roots of x² + x + 1 = 0

$x^3-1=(x-1)(x^2+x+1)$, so $\alpha^3=\beta^3=1$. Also $\alpha+1=-\alpha^2$, $\alpha\beta=1$, $\alpha+\beta=-1$.

Step 2. (P)

$$\frac1{(\alpha+1)^{2026}}=\frac1{\alpha^{4052}}=\frac1{\alpha^2}=\alpha$$

since $4052=3\cdot1350+2$ and $\alpha^3=1$. Likewise for $\beta$. The roots are $\alpha,\beta$ again: equation (1).

Step 3. (Q)

The odd power adds a minus sign: the roots become $-\alpha^{-4054}=-\alpha^{-1}=-\alpha^2$ and $-\beta^2$. Sum $-(\alpha^2+\beta^2)=-(1-2)=1$, product $1$: equation (2).

Step 4. (R)

$x^3+1=(x+1)(x^2-x+1)$ gives $\gamma^3=-1$, so $\gamma^6=1$, and $\gamma-1=\gamma^2$:

$$\frac1{\gamma^{4052}}+\frac1{\delta^{4052}}=\frac1{\gamma^2}+\frac1{\delta^2}$$$$=\frac{(\gamma+\delta)^2-2\gamma\delta}{(\gamma\delta)^2}=\frac{1-2}{1}=-1$$

That is (4).

Step 5. (S)

Put $u=p+1$, $v=r+1$: $u+v=1$, $uv=pr+p+r+1=-1$.

$$\frac{u^3+v^3}{(uv)^3}=\frac{1+3}{-1}=-4$$

That is (5).

Answer: option (3)

Power sums and Newton's formula

Let $\alpha,\beta$ be the roots of $ax^2+bx+c=0$ and $S_k=\alpha^k+\beta^k$. Multiply $a\alpha^2+b\alpha+c=0$ by $\alpha^{k-2}$, do the same for $\beta$, and add:

$$aS_k+bS_{k-1}+cS_{k-2}=0,\qquad k\ge2$$

This is Newton's formula. Start from $S_0=2$ and $S_1=-\tfrac ba$ and step up one power at a time. The same proof shows that $T_k=\alpha^k-\beta^k$, and any $p\alpha^k+q\beta^k$, satisfy the same recurrence.

Example 25 (IITianForum). If $\alpha,\beta$ are the roots of $x^2-3x+1=0$, find $\alpha^5+\beta^5$.

▸ SHOW SOLUTION

Step 1. Recurrence

$S_k=3S_{k-1}-S_{k-2}$, with $S_0=2$, $S_1=3$.

Step 2. Step up

$$S_2=7,\ S_3=18,\ S_4=47,\ S_5=123$$

Answer: $123$

Example 26 (IIT JEE 2011). Let $\alpha,\beta$ be the roots of $x^2-6x-2=0$, with $\alpha>\beta$. If $a_n=\alpha^n-\beta^n$ for $n\ge1$, then the value of

$$\frac{a_{10}-2a_8}{2a_9}$$

is (A) $1$ (B) $2$ (C) $3$ (D) $4$

▸ SHOW SOLUTION

Step 1. Recurrence

$\alpha^2=6\alpha+2$ and $\beta^2=6\beta+2$. Multiply by $\alpha^8$, $\beta^8$ and subtract:

$$a_{10}=6a_9+2a_8$$

Step 2. Value

$$\frac{a_{10}-2a_8}{2a_9}=\frac{6a_9}{2a_9}=3$$

Answer: option (C)

Example 27 (JEE Main 2025, 3 April Shift 1). Let $\alpha,\beta$ be the roots of $x^2+\sqrt3x-16=0$, and $\gamma,\delta$ the roots of $x^2+3x-1=0$. If $P_n=\alpha^n+\beta^n$ and $Q_n=\gamma^n+\delta^n$, then

$$\frac{P_{25}+\sqrt3P_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}$$

is equal to (1) $3$ (2) $4$ (3) $5$ (4) $7$

▸ SHOW SOLUTION

Step 1. First part

$P_{25}+\sqrt3P_{24}-16P_{23}=0$, so the first fraction is $\tfrac{16}2=8$.

Step 2. Second part

$Q_{25}+3Q_{24}-Q_{23}=0$, so $Q_{25}-Q_{23}=-3Q_{24}$ and the second fraction is $-3$.

Answer: $8-3=5$, option (3)

Example 28 (JEE Main 2025, 2 April Shift 1). Let $P_n=\alpha^n+\beta^n$. If $P_{10}=123$, $P_9=76$, $P_8=47$ and $P_1=1$, then the quadratic equation with roots $\tfrac1\alpha$ and $\tfrac1\beta$ is

▸ SHOW SOLUTION

Step 1. Find the equation of α, β

Let $\alpha,\beta$ be roots of $x^2-sx+p=0$, so $P_{10}=sP_9-pP_8$. Here $s=P_1=1$:

$$123=76-47p,\qquad p=-1$$

So $\alpha,\beta$ are the roots of $x^2-x-1=0$.

Step 2. Reciprocals

Reverse the coefficients: $-x^2-x+1=0$.

Answer: $x^2+x-1=0$, option (2)

Example 29 (JEE Main 2026, 6 April Shift 2). Let $S=\{z\in\mathbb C:z^2+\sqrt6\,iz-3=0\}$. Then $\sum_{z\in S}z^8$ is equal to

▸ SHOW SOLUTION

Step 1. Sum and product

For the roots $\alpha,\beta$: $\alpha+\beta=-\sqrt6\,i$, $\alpha\beta=-3$.

$$\alpha^2+\beta^2=(-\sqrt6\,i)^2+6=-6+6=0$$

Step 2. Square twice

$$\alpha^4+\beta^4=0-2(\alpha\beta)^2=-18$$$$\alpha^8+\beta^8=(-18)^2-2(\alpha\beta)^4$$$$=324-162=162$$

Answer: $162$, option (1)

Example 30 (JEE Advanced 2017). Let $p,q$ be integers and $\alpha,\beta$ the roots of $x^2-x-1=0$, $\alpha\ne\beta$. For $n=0,1,2,\dots$ let $a_n=p\alpha^n+q\beta^n$. (Fact: if $a,b$ are rational and $a+b\sqrt5=0$, then $a=b=0$.) (i) Express $a_{12}$ in terms of $a_{11}$ and $a_{10}$. (ii) If $a_4=28$, find $p+2q$.

▸ SHOW SOLUTION

Step 1. Recurrence

$\alpha^2=\alpha+1$, $\beta^2=\beta+1$, so $a_{n+2}=a_{n+1}+a_n$. In particular $a_{12}=a_{11}+a_{10}$.

Step 2. a₄ in p and q

$a_2=a_1+a_0$, $a_3=2a_1+a_0$, $a_4=3a_1+2a_0$. With $a_0=p+q$ and $\alpha,\beta=\tfrac{1\pm\sqrt5}2$:

$$a_4=\frac72(p+q)+\frac{3\sqrt5}{2}(p-q)=28$$

Step 3. Use the fact

$p-q=0$ and $\tfrac72(p+q)=28$, so $p=q=4$.

Answer: (i) $a_{12}=a_{11}+a_{10}$; (ii) $p+2q=12$

Example 31 (IITianForum, JEE Advanced level). Let $\alpha>\beta$ be the roots of $x^2-x-1=0$. For $n\ge1$ let

$$a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta},\qquad b_n=\alpha^n+\beta^n$$

Which are true? (A) $a_1+a_2+\cdots+a_n=a_{n+2}-1$ (B) $\sum_{n=1}^{\infty}\tfrac{a_n}{10^n}=\tfrac{10}{89}$ (C) $\sum_{n=1}^{\infty}\tfrac{b_n}{10^n}=\tfrac8{89}$ (D) $b_n=a_{n-1}+a_{n+1}$ for $n\ge2$

▸ SHOW SOLUTION

Step 1. Recurrences

Both sequences satisfy $u_{n+2}=u_{n+1}+u_n$, with $a_1=a_2=1$ and $b_1=1$, $b_2=3$.

Step 2. (A)

$a_k=a_{k+2}-a_{k+1}$. Adding for $k=1$ to $n$, the sum telescopes to $a_{n+2}-a_2=a_{n+2}-1$. True.

Step 3. (B) and (C)

Let $A=\sum\frac{a_n}{10^n}$. The recurrence gives

$$A-\frac{A}{10}-\frac{A}{100}=\frac{a_1}{10}+\frac{a_2-a_1}{100}=\frac1{10}$$

so $A=\tfrac{10}{89}$. True. The same steps with $b_1=1$, $b_2=3$ give

$$B\cdot\frac{89}{100}=\frac1{10}+\frac2{100},\qquad B=\frac{12}{89}$$

So (C) is false.

Step 4. (D)

$a_{n-1}+a_{n+1}=\dfrac{\alpha^{n-1}(1+\alpha^2)-\beta^{n-1}(1+\beta^2)}{\alpha-\beta}$. From $\alpha\beta=-1$, $1+\alpha^2=\alpha(\alpha-\beta)$, and $1+\beta^2=-\beta(\alpha-\beta)$. So the sum is $\alpha^n+\beta^n=b_n$. True.

Answer: (A), (B), (D)

Graph of the quadratic expression

Completing the square,

$$y=a\left(x+\frac b{2a}\right)^2-\frac{D}{4a}$$

So the graph of $y=ax^2+bx+c$ is a parabola with

$$\text{vertex}\ \left(-\frac b{2a},\,-\frac D{4a}\right)$$ $$\text{axis}\ x=-\frac b{2a}$$

• It opens upwards for $a>0$ and downwards for $a<0$.
• It meets the $y$-axis at $(0,c)$.
• It meets the $x$-axis at the real roots: two points if $D>0$, touches at one point if $D=0$, misses if $D<0$.

A parabola opening upwards with its vertex below the x axis marked, a dashed vertical axis of symmetry through the vertex, the two points where it cuts the x axis marked alpha and beta, and the point where it cuts the y axis marked 0 comma c

The sign of $a$ and the sign of $D$ give the six pictures. Nearly every question on sign, range or location of roots is one of these.

Six small graphs in two rows; top row parabolas opening upwards, bottom row opening downwards; in each row the left one cuts the x axis at two marked points, the middle one touches it at one marked point, and the right one does not meet it

Top row $a>0$, bottom row $a<0$; left to right $D>0$, $D=0$, $D<0$.

Reading signs from a graph. $a$ from the opening, $c$ from the $y$-intercept, and $b$ from the vertex: $-\tfrac b{2a}>0$ (vertex right of the $y$-axis) means $a,b$ have opposite signs.

Example 32 (IITianForum). (a) Sketch $y=-2x^2+4x+6$, marking the vertex, the roots and the $y$-intercept. (b) A parabola $y=ax^2+bx+c$ opens downwards, has its vertex to the right of the $y$-axis and meets the $y$-axis below the origin. Find the signs of $a$, $b$, $c$ and $D$, given that it meets the $x$-axis at two points.

▸ SHOW SOLUTION

Step 1. Vertex form

$$y=-2(x-1)^2+8$$

Vertex $(1,8)$, opening downwards.

Step 2. Roots and intercept

$-2x^2+4x+6=0$ gives $x^2-2x-3=0$, roots $-1$ and $3$. The $y$-intercept is $6$.

Step 3. Part (b)

Downwards: $a<0$. Vertex right: $-\tfrac b{2a}>0$, so $b>0$. Below the origin: $c<0$. Two meeting points: $D>0$.

Answer: (b) $a<0$, $b>0$, $c<0$, $D>0$

A downward parabola with its vertex marked at 1 comma 8, a dashed line from the vertex down to the x axis, the points minus 1 and 3 marked where it cuts the x axis, and the point 6 marked on the y axis

Sign of the quadratic expression

Let $f(x)=ax^2+bx+c$, $a,b,c$ real, $a\ne0$.

Reason. For $D>0$, $f(x)=a(x-\alpha)(x-\beta)$. For $x$ between the roots the two brackets have opposite signs; outside, the same sign. For $D<0$, the vertex form is $a$ times a positive number.

An upward parabola cutting the x axis at alpha and beta, with the part between them shaded below the axis and a minus sign there, and plus signs above the axis to the left of alpha and to the right of beta

$$ax^2+bx+c>0\ \forall x\iff a>0,\ D<0$$ $$ax^2+bx+c<0\ \forall x\iff a<0,\ D<0$$ $$ax^2+bx+c\ge0\ \forall x\iff a>0,\ D\le0$$

Quadratic inequalities. Find the roots, then read the sign: for $a>0$, $f(x)<0$ exactly between the roots and $f(x)>0$ outside them. For products and quotients of several factors, use the wavy curve method of the Basic Algebra and Logarithms note; a quadratic factor with $D<0$ and positive leading coefficient can simply be dropped, since it is always positive.

Example 33 (IITianForum). Find all real $k$ for which $kx^2+4x+k>0$ for every real $x$.

▸ SHOW SOLUTION

Step 1. k = 0

The expression is $4x$, not always positive.

Step 2. k ≠ 0

Need $k>0$ and $D=16-4k^2<0$, that is $k^2>4$.

Answer: $k>2$

Example 34 (IITianForum). Let $f(x)=ax^2+bx+c$ with $a<b$ and $f(x)\ge0$ for every real $x$. Find the least value of

$$\frac{4a+2b+c}{b-a}$$

▸ SHOW SOLUTION

Step 1. Conditions

$f\ge0$ for all $x$ forces $a>0$ and $b^2\le4ac$, so $c\ge\tfrac{b^2}{4a}$. The expression is unchanged when $a,b,c$ are scaled, so take $a=1$ and $b>1$:

$$\frac{4+2b+c}{b-1}\ge\frac{4+2b+\frac{b^2}4}{b-1}=\frac{(b+4)^2}{4(b-1)}$$

Step 2. Substitute and use AM ≥ GM

Put $u=b-1>0$:

$$\frac{(u+5)^2}{4u}\ge\frac{(2\sqrt{5u})^2}{4u}=5$$

Equality at $u=5$: $b=6$, $c=9$, $f(x)=(x+3)^2$.

Answer: $5$

Maximum, minimum and range

From the vertex form, $f(x)=ax^2+bx+c$ has its extreme value at $x=-\tfrac b{2a}$:

$$f\left(-\frac b{2a}\right)=-\frac D{4a}=\frac{4ac-b^2}{4a}$$

On a closed interval $[p,q]$. Compare $f(p)$, $f(q)$ and, if the vertex lies in $[p,q]$, $f\left(-\tfrac b{2a}\right)$. For $a>0$: the minimum is at the vertex when it lies inside, otherwise at the end nearer the vertex; the maximum is always at the end farther from the vertex.

Example 35 (IITianForum). Find the maximum and minimum values of $f(x)=x^2-4x+1$ on (a) $[-1,3]$; (b) $[3,5]$.

▸ SHOW SOLUTION

Step 1. Vertex

$f(x)=(x-2)^2-3$: vertex at $x=2$, value $-3$.

Step 2. Interval [−1, 3]

The vertex lies inside: minimum $-3$. Ends: $f(-1)=6$, $f(3)=-2$; maximum $6$.

Step 3. Interval [3, 5]

The vertex lies to the left, so $f$ increases here: minimum $f(3)=-2$, maximum $f(5)=6$.

Answer: (a) max $6$, min $-3$; (b) max $6$, min $-2$

Two copies of the parabola y equals x squared minus 4x plus 1 drawn dashed; on the left the part over the interval from minus 1 to 3 is drawn solid with the end points minus 1 comma 6 and 3 comma minus 2 and the vertex 2 comma minus 3 marked; on the right the part over the interval from 3 to 5 is drawn solid with the end points 3 comma minus 2 and 5 comma 6 marked

Range of a rational expression

To find the values of

$$y=\frac{a_1x^2+b_1x+c_1}{a_2x^2+b_2x+c_2}$$

clear the denominator and read the result as a quadratic in $x$:

$$(a_1-ya_2)x^2+(b_1-yb_2)x$$ $$+(c_1-yc_2)=0$$

A value $y$ is taken exactly when this has a real root $x$ that does not make the denominator zero. So the range is given by

$$(b_1-yb_2)^2-4(a_1-ya_2)(c_1-yc_2)\ge0$$

together with two checks.

1. The linear case. For the $y$ with $a_1-ya_2=0$ the equation is linear; solve it directly.
2. Common factor. If the numerator and denominator share a factor $(x-r)$, cancel it first. The value of the cancelled form at $x=r$ is not taken, although the discriminant test lets it through.

If the two quadratics have no common factor, the test is correct even when the denominator has real roots.

Example 36 (JEE Main 2026, 6 April Shift 2). Let $f:\mathbb R\to\mathbb R$, $f(x)=\dfrac{2x^2-3x+2}{3x^2+x+3}$. Then $f$ is

▸ SHOW SOLUTION

Step 1. Quadratic in x

$$(3y-2)x^2+(y+3)x+(3y-2)=0$$

The denominator has $D=1-36<0$, so it never vanishes.

Step 2. Range

$$(y+3)^2-4(3y-2)^2\ge0$$$$(7y-1)(5y-7)\le0,\qquad \tfrac17\le y\le\tfrac75$$

The range is $\left[\tfrac17,\tfrac75\right]\ne\mathbb R$: not onto.

Step 3. Not one-one

The equation in $x$ has its coefficients of $x^2$ and of $1$ equal, so its two roots have product $1$: if $x$ gives the value $y$, so does $\tfrac1x$. For example $f(2)=f\left(\tfrac12\right)=\tfrac4{17}$.

Answer: option (4)

The curve y equals 2x squared minus 3x plus 2 over 3x squared plus x plus 3, rising to a highest point at x equals minus 1 on the dashed line y equals 7 over 5, falling to a lowest point at x equals 1 on the dashed line y equals 1 over 7, and levelling off towards the dotted line y equals 2 over 3 at both ends

Example 37 (IITianForum). Exactly one integer is not in the range of $f(x)=\dfrac{x^2-5x+6}{x^2+x-12}$. Find it, and find the full range.

▸ SHOW SOLUTION

Step 1. Cancel the common factor

$$f(x)=\frac{(x-2)(x-3)}{(x+4)(x-3)}=\frac{x-2}{x+4}$$

for $x\ne3,\ -4$.

Step 2. Values of the cancelled form

$\dfrac{x-2}{x+4}=1-\dfrac6{x+4}$ takes every real value except $1$.

Step 3. Remove x = 3

At $x=3$ the cancelled form equals $\tfrac17$, and no other $x$ gives $\tfrac17$. So $\tfrac17$ is not taken either.

$$\text{Range}=\mathbb R-\left\{1,\tfrac17\right\}$$

Answer: the integer is $1$

The discriminant test alone would miss $\tfrac17$ here: it gives $(7y-1)^2\ge0$, true for every $y$, and at $y=\tfrac17$ it produces the double root $x=3$, which is not in the domain.

The curve y equals x minus 2 over x plus 4 in two branches with dashed asymptotes x equals minus 4 and y equals 1, and a hollow circle on the right branch at the point 3 comma 1 over 7 showing a missing value

Example 38 (IITianForum). $P(x)$ is a quadratic polynomial with real coefficients such that $2(1+P(x))=P(x-1)+P(x+1)$ for every real $x$, with $P(0)=5$ and $P(3)=29$. Find (a) the sum of the coefficients of $P$; (b) the range of $P$; (c) $P(10)$.

▸ SHOW SOLUTION

Step 1. Leading coefficient

For $P(x)=Ax^2+Bx+C$:

$$P(x-1)+P(x+1)=2P(x)+2A$$

So $2+2P(x)=2P(x)+2A$, giving $A=1$.

Step 2. The polynomial

$C=P(0)=5$ and $9+3B+5=29$, so $B=5$:

$$P(x)=x^2+5x+5=\left(x+\tfrac52\right)^2-\tfrac54$$

Step 3. The three answers

Sum of coefficients $P(1)=11$. Range $\left[-\tfrac54,\infty\right)$. $P(10)=155$.

Answer: (a) $11$; (b) $\left[-\tfrac54,\infty\right)$; (c) $155$

Location of roots

Let $f(x)=ax^2+bx+c$ with $a>0$ (if $a<0$, multiply the equation by $-1$), real roots $\alpha\le\beta$, and real numbers $k$, $k_1<k_2$. Each condition is the parabola picture written down: the sign of $f$ at a test point tells which side of the roots that point is on, the vertex $-\tfrac b{2a}$ tells where the turning point is, and $D$ decides whether real roots exist.

For any sign of $a$, "$k$ between the roots" is $af(k)<0$, and "$f(k)>0$" in the other rows becomes $af(k)>0$.

Three upward parabolas; in the first a point k on the x axis lies to the left of both roots with f of k positive; in the second k lies to the right of both roots with f of k positive; in the third k lies between the roots with f of k negative; a dashed line marks the vertex in each

Two upward parabolas with an interval from k1 to k2 highlighted on the x axis; on the left f of k1 is negative and f of k2 positive, so exactly one root lies in the interval; on the right f of k1 and f of k2 are both positive and both roots lie inside the interval

Example 39 (IITianForum). Find all real $k$ for which both roots of $x^2-2kx+k^2+k-5=0$ are less than $5$.

▸ SHOW SOLUTION

Step 1. Real roots

$$\frac D4=k^2-(k^2+k-5)=5-k$$

$D\ge0$ gives $k\le5$.

Step 2. f(5) > 0

$$f(5)=k^2-9k+20$$$$=(k-4)(k-5)>0$$

So $k<4$ or $k>5$.

Step 3. Vertex

$-\tfrac b{2a}=k<5$.

Answer: $k<4$

Example 40 (JEE Main 2026, 21 January Shift 2). Let $\alpha,\beta$ be the roots of $x^2+2ax+(3a+10)=0$ such that $\alpha<1<\beta$. Then the set of all possible values of $a$ is

▸ SHOW SOLUTION

Step 1. One condition

The parabola opens upwards and $1$ lies between the roots, so $f(1)<0$.

Step 2. Solve

$$1+2a+3a+10<0,\qquad a<-\frac{11}5$$

Answer: option (4)

Example 41 (JEE Main 2026, 4 April Shift 2). If $(\lambda+2)x^2-3\lambda x+4\lambda=0$, $\lambda\ne-2$, has two positive roots, then the number of possible integral values of $\lambda$ is

▸ SHOW SOLUTION

Step 1. Product and sum positive

$$\frac{4\lambda}{\lambda+2}>0,\qquad \frac{3\lambda}{\lambda+2}>0$$

Both give $\lambda<-2$ or $\lambda>0$.

Step 2. Real roots

$$D=9\lambda^2-16\lambda(\lambda+2)$$$$=-\lambda(7\lambda+32)\ge0$$

So $-\tfrac{32}7\le\lambda\le0$.

Step 3. Combine

$\lambda\in\left[-\tfrac{32}7,-2\right)$: integers $-4,-3$.

Answer: $2$, option (2)

Example 42 (JEE Main 2025, 7 April Shift 1). Let the set of all $p\in\mathbb R$ for which both roots of $x^2-(p+2)x+(2p+9)=0$ are negative real numbers be $(\alpha,\beta]$. Then $\beta-2\alpha$ is

▸ SHOW SOLUTION

Step 1. Real roots

$$(p+2)^2-4(2p+9)\ge0$$$$(p+4)(p-8)\ge0$$

So $p\le-4$ or $p\ge8$.

Step 2. Both negative

Sum $p+2<0$ and product $2p+9>0$: $-\tfrac92<p<-2$.

Step 3. Combine

$p\in\left(-\tfrac92,-4\right]$.

Answer: $\beta-2\alpha=-4+9=5$, option (3)

Example 43 (JEE Main 2026, 28 January Shift 2). Let the arithmetic mean of $\tfrac1a$ and $\tfrac1b$ be $\tfrac5{16}$, $a>2$. If $a,4,\alpha,b$ are in A.P., then the equation $\alpha x^2-ax+2(\alpha-2b)=0$ has

▸ SHOW SOLUTION

Step 1. The A.P.

With common difference $d$: $a=4-d$, $\alpha=4+d$, $b=4+2d$.

$$\frac1{4-d}+\frac1{4+2d}=\frac58$$$$5d^2-6d-8=0,\qquad d=2\ \text{or}\ -\tfrac45$$

Both give $a>2$.

Step 2. The equation

$d=2$: $a=2$, $\alpha=6$, $b=8$, and the equation is $3x^2-x-10=0$, roots $2$ and $-\tfrac53$. $d=-\tfrac45$ gives $2x^2-3x-2=0$, roots $2$ and $-\tfrac12$.

Step 3. Location

In both cases one root is $2\in(1,4)$ and the other lies in $(-2,0)$.

Answer: option (1)

Example 44 (IITianForum). The equation $(c-3)x^2-2cx+(c-2)=0$ has both roots positive, one root less than $2$ and the other between $2$ and $3$. Find the number of integer values of $c$.

▸ SHOW SOLUTION

Step 1. Test points

With $A=c-3$ the leading coefficient: $2$ lies between the roots, $0$ and $3$ lie outside them. So

$$A f(2)<0,\quad A f(0)>0,\quad A f(3)>0$$

Step 2. Values

$$f(2)=c-14,\quad f(0)=c-2$$$$f(3)=4c-29$$

Step 3. Solve

$(c-3)(c-14)<0$: $3<c<14$. Then $c-3>0$, so $f(0)>0$ holds and $f(3)>0$ needs $c>\tfrac{29}4$.

$c\in\left(\tfrac{29}4,14\right)$: integers $8$ to $13$.

Answer: $6$

Common roots

Take $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$.

One common root. Let $\alpha$ be common. Treat $\alpha^2$ and $\alpha$ as unknowns and solve the two linear equations (cross-multiplication):

$$\frac{\alpha^2}{b_1c_2-b_2c_1}=\frac{\alpha}{c_1a_2-c_2a_1}=\frac1{a_1b_2-a_2b_1}$$

So

$$\alpha=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}$$

and the condition for a common root is

$$(c_1a_2-c_2a_1)^2$$ $$=(b_1c_2-b_2c_1)(a_1b_2-a_2b_1)$$

In practice it is often quicker to subtract a suitable multiple of one equation from the other to remove $x^2$, find $\alpha$, and substitute back.

Both roots common. The equations must be multiples of each other:

$$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$ The parabolas y equals x squared minus 2x plus 1, touching the x axis at 1, and y equals x squared plus x minus 2, cutting the x axis at minus 2 and 1, meeting each other at the common root 1 on the x axis

Example 45 (IITianForum). Find $k$ if $x^2+kx+1=0$ and $x^2+x+k=0$ have a common root.

▸ SHOW SOLUTION

Step 1. Subtract

$$(k-1)x+(1-k)=0$$

Step 2. Two cases

$k\ne1$: the common root is $x=1$, and $1+k+1=0$ gives $k=-2$. Then the equations are $(x-1)^2=0$ and $(x-1)(x+2)=0$.

$k=1$: both equations are $x^2+x+1=0$, with both (non-real) roots common.

Answer: $k=-2$ for one common real root; $k=1$ makes the equations identical

Example 46 (IITianForum). Consider $x^2-2p_rx+c_r=0$ for $r=1,2,3$ with $c_1=2$, $c_2=3$, $c_3=6$. Every pair of these equations has exactly one root in common, and no number is a root of all three. (a) Find the root common to the first and the third equation. (b) How many triplets $(p_1,p_2,p_3)$ are possible?

▸ SHOW SOLUTION

Step 1. Name the roots

The equations share roots in a cycle: first $\{u,v\}$, second $\{u,w\}$, third $\{v,w\}$. Products:

$$uv=2,\qquad uw=3,\qquad vw=6$$

Step 2. Solve

$(uvw)^2=36$, so $uvw=\pm6$. Then $w=\pm3$, $v=\pm2$, $u=\pm1$, all with the same sign.

Step 3. Answers

The first and third share $v=\pm2$. The sign is the only freedom, and each sign fixes $2p_1=u+v$, $2p_2=u+w$, $2p_3=v+w$.

Answer: (a) $2$ or $-2$; (b) $2$

Equations reducible to a quadratic

For the reciprocal form use $x^2+\tfrac1{x^2}=t^2-2$.

Example 47 (IITianForum). Solve $x^4-5x^3+8x^2-5x+1=0$.

▸ SHOW SOLUTION

Step 1. Divide by x²

$$x^2+\frac1{x^2}-5\left(x+\frac1x\right)+8=0$$

Step 2. t = x + 1/x

$$t^2-5t+6=0,\qquad t=2,3$$

Step 3. Back to x

$x+\tfrac1x=2$ gives $x=1$ (twice). $x+\tfrac1x=3$ gives $x^2-3x+1=0$.

Answer: $x=1,\ \dfrac{3\pm\sqrt5}2$

Example 48 (JEE Main 2026, 24 January Shift 2). The smallest positive integral value of $a$ for which all the roots of $x^4-ax^2+9=0$ are real and distinct is

▸ SHOW SOLUTION

Step 1. Quadratic in t = x²

$t^2-at+9=0$ needs two distinct positive roots; each gives two values $\pm\sqrt t$.

Step 2. Conditions

$D=a^2-36>0$, sum $a>0$, product $9>0$. So $a>6$.

Answer: $a=7$, option (4)

Example 49 (JEE Main 2026, 23 January Shift 1). If $\alpha<\beta$ are the roots of

$$(\sqrt3-2)\lvert\sqrt x-3\rvert+(x-6\sqrt x)$$ $$+(9-2\sqrt3)=0,\quad x\ge0$$

then $\sqrt{\tfrac\beta\alpha}+\sqrt{\alpha\beta}$ is

▸ SHOW SOLUTION

Step 1. Substitute u = |√x − 3|

$x-6\sqrt x=u^2-9$:

$$u^2+(\sqrt3-2)u-2\sqrt3=0$$$$(u+\sqrt3)(u-2)=0$$

$u\ge0$, so $u=2$.

Step 2. Back to x

$\sqrt x=5$ or $1$: $x=25$ or $1$.

Answer: $\sqrt{25}+\sqrt{25}=10$, option (3)

Example 50 (IITianForum). Let $b>0$, $b\ne4$. Show that $x^2-b|x|+4=0$ has either four real roots or no real root.

▸ SHOW SOLUTION

Step 1. t = |x|

$$t^2-bt+4=0$$

Product $4>0$ and sum $b>0$: when real, both roots in $t$ are positive.

Step 2. Two cases

$b>4$: $D=b^2-16>0$, two distinct positive $t$, each giving $x=\pm t$: four roots. $0<b<4$: $D<0$, no real $t$, no real $x$.

Answer: four roots for $b>4$, none for $0<b<4$; two roots ($\pm2$) only for the excluded $b=4$

Two graphs; on the left y equals x squared minus 5 mod x plus 4, a W shaped curve symmetric about the y axis cutting the x axis at minus 4, minus 1, 1 and 4; on the right y equals x squared minus 3 mod x plus 4, the same shape lying wholly above the x axis

Example 51 (IITianForum). Let $\alpha,\beta$ be the real roots of $(\log_2x)^2-8\log_2x+k=0$, $k\in\mathbb R$. Find the least possible value of $\alpha+\beta$.

▸ SHOW SOLUTION

Step 1. t = log₂x

$t^2-8t+k=0$ with $t_1+t_2=8$, so

$$\alpha\beta=2^{t_1+t_2}=2^8=256$$

Step 2. AM ≥ GM

$$\alpha+\beta\ge2\sqrt{256}=32$$

Equality needs $\alpha=\beta=16$, that is $t_1=t_2=4$, which happens for $k=16$.

Answer: $32$

Quadratic expression in two variables

The general second degree expression in $x$ and $y$ is

$$ax^2+2hxy+by^2+2gx+2fy+c$$

It splits into two linear factors exactly when

$$\Delta=abc+2fgh-af^2-bg^2-ch^2=0$$

Reason ($a\ne0$). Read it as a quadratic in $x$:

$$ax^2+2(hy+g)x+(by^2+2fy+c)$$

It factors into linear factors in $x$ and $y$ exactly when its discriminant is a perfect square in $y$. A quarter of that discriminant is

$$(h^2-ab)y^2+2(gh-af)y+(g^2-ac)$$

and a quadratic in $y$ is a perfect square when its own discriminant is zero:

$$(gh-af)^2=(h^2-ab)(g^2-ac)$$

Expanding and dividing by $a$ gives $\Delta=0$.

Range of one variable. If real $x,y$ satisfy a quadratic relation, read it as a quadratic in $y$ (or $x$) and demand $D\ge0$. For example $x^2-xy+y^2=3$ read in $y$ gives $x^2-4(x^2-3)\ge0$, so $-2\le x\le2$.

Example 52 (IITianForum). Find $k$ so that $x^2-5xy+6y^2+x-4y+k$ is a product of two linear factors, and find the factors.

▸ SHOW SOLUTION

Step 1. Read off a, b, h, g, f

$a=1$, $b=6$, $h=-\tfrac52$, $g=\tfrac12$, $f=-2$, $c=k$.

Step 2. The condition

$$6k+2(-2)\left(\tfrac12\right)\left(-\tfrac52\right)$$$$-4-6\cdot\tfrac14-\tfrac{25}4k=0$$$$-\tfrac14k-\tfrac12=0,\qquad k=-2$$

Step 3. Factors

$x^2-5xy+6y^2=(x-2y)(x-3y)$. Try $(x-2y+p)(x-3y+q)$: $p+q=1$ and $3p+2q=4$ give $p=2$, $q=-1$, and $pq=-2=k$.

Answer: $k=-2$; $(x-2y+2)(x-3y-1)$

Polynomial equations of higher degree

For $a_0x^n+a_1x^{n-1}+\cdots+a_n=0$, $a_0\ne0$, with roots $\alpha_1,\dots,\alpha_n$, compare $a_0(x-\alpha_1)\cdots(x-\alpha_n)$ with the polynomial:

$$\sum\alpha_1=-\frac{a_1}{a_0},\quad \sum\alpha_1\alpha_2=\frac{a_2}{a_0}$$ $$\alpha_1\alpha_2\cdots\alpha_n=(-1)^n\frac{a_n}{a_0}$$

The cubic with roots $\alpha,\beta,\gamma$ is

$$x^3-\left(\textstyle\sum\alpha\right)x^2+\left(\textstyle\sum\alpha\beta\right)x-\alpha\beta\gamma=0$$

Facts about the roots.

1. A polynomial equation of degree $n$ has exactly $n$ roots, counted with repetition.
2. With real coefficients, non-real roots come in conjugate pairs; so an equation of odd degree with real coefficients has at least one real root.
3. With rational coefficients, irrational roots $p+\sqrt q$ come with $p-\sqrt q$.
4. Remainder theorem: $f(x)$ divided by $x-a$ leaves $f(a)$. Factor theorem: $x-a$ is a factor exactly when $f(a)=0$.
5. If $f(p)$ and $f(q)$ have opposite signs ($f$ a polynomial), a real root lies between $p$ and $q$.
6. Descartes' rule of signs: the number of positive roots is at most the number of sign changes in the coefficients of $f(x)$, and the number of negative roots at most the number of sign changes in $f(-x)$; in each case the shortfall is even.

Two graphs; on the left the cubic y equals x cubed minus x squared plus 1 coming up from below the x axis and crossing it once at a marked point; on the right the quartic y equals x to the fourth minus 2x squared plus 2, a W shaped curve that stays above the x axis

Newton's formula for a cubic. For $ax^3+bx^2+cx+d=0$ with $S_k=\sum\alpha^k$:

$$aS_1+b=0$$ $$aS_2+bS_1+2c=0$$ $$aS_3+bS_2+cS_1+3d=0$$ $$aS_k+bS_{k-1}+cS_{k-2}+dS_{k-3}=0$$

for $k\ge4$.

Example 53 (IITianForum). Numbers $\alpha,\beta,\gamma$ satisfy $\alpha+\beta+\gamma=3$, $\alpha^2+\beta^2+\gamma^2=5$ and $\alpha^3+\beta^3+\gamma^3=9$. Find $\alpha^4+\beta^4+\gamma^4$.

▸ SHOW SOLUTION

Step 1. The cubic

They are the roots of $x^3-e_1x^2+e_2x-e_3=0$ with $e_1=3$ and

$$e_2=\frac{e_1^2-S_2}{2}=\frac{9-5}2=2$$

Step 2. e₃ from S₃

$S_3-e_1S_2+e_2S_1-3e_3=0$:

$$9-15+6-3e_3=0,\qquad e_3=0$$

Step 3. S₄

$$S_4=e_1S_3-e_2S_2+e_3S_1$$$$=27-10+0=17$$

(Here the cubic is $x(x-1)(x-2)$, roots $0,1,2$: $0+1+16=17$.)

Answer: $17$

Example 54 (IITianForum). Let $\alpha_1,\dots,\alpha_5$ be the roots of $x^5-3x^4-1=0$. Find $\displaystyle\sum_{r=1}^5\frac1{\alpha_r^4}$ and $\displaystyle\prod_{r=1}^5\left(\frac1{\alpha_r^4}+3\right)$.

▸ SHOW SOLUTION

Step 1. Rewrite 1/α⁴

$\alpha^5=3\alpha^4+1$. Divide by $\alpha^4$:

$$\frac1{\alpha^4}=\alpha-3$$

Step 2. Sum

$\sum\alpha_r=3$:

$$\sum\frac1{\alpha_r^4}=3-15=-12$$

Step 3. Product

$\tfrac1{\alpha^4}+3=\alpha$, and the product of the roots of $x^5-3x^4-1=0$ is $(-1)^5\cdot(-1)=1$.

Answer: $-12$ and $1$

Example 55 (JEE Advanced 2025, Paper 1). Let $a_i,b_i\in\mathbb R$ for $i\in\{1,2,3\}$. Define $f,g,h:\mathbb R\to\mathbb R$ by

$$f(x)=a_1+10x+a_2x^2+a_3x^3+x^4$$ $$g(x)=b_1+3x+b_2x^2+b_3x^3+x^4$$

and $h(x)=f(x+1)-g(x+2)$. If $f(x)\ne g(x)$ for every $x\in\mathbb R$, then the coefficient of $x^3$ in $h(x)$ is

▸ SHOW SOLUTION

Step 1. f − g has no real root

$$f(x)-g(x)=(a_3-b_3)x^3+(a_2-b_2)x^2$$$$+7x+(a_1-b_1)$$

If $a_3\ne b_3$ this has odd degree $3$ and so a real root, against $f(x)\ne g(x)$. Hence $a_3=b_3$.

Step 2. Coefficient of x³ in h

From $f(x+1)$: $a_3+4$. From $g(x+2)$: $b_3+4\cdot2=b_3+8$.

$$a_3+4-b_3-8=-4$$

Answer: option (C)

The same reasoning shows why the remaining quadratic $(a_2-b_2)x^2+7x+(a_1-b_1)$ must have $D<0$: it may never vanish, so it keeps one sign.

Which method to use

A decision chart: from the question asked, arrows lead to eight situations, each paired with its method: nature of roots, the sign of D or a perfect square; expression in the roots, write it in S and P; high powers of roots, Newton's formula; new equation, substitute or use the new sum and product; sign, maximum, minimum or range, the vertex and the six pictures; roots on one side of k, D, the sign of a f of k and the vertex; common root, subtract or cross-multiply; modulus, exponential or reciprocal equations, substitute t and keep the allowed values of t

What JEE asks

Across the 2025 and 2026 JEE Main papers and recent JEE Advanced papers the chapter appears in these forms.

1. Relations between roots and coefficients, usually with a symmetric expression or a difference of roots (Examples 10 to 19). This is the most common form in JEE Main.
2. Location and sign of roots with a parameter (Examples 39 to 44).
3. Power sums by Newton's recurrence, often with large powers or complex roots (Examples 26 to 31).
4. Equations reducible to a quadratic: biquadratic, modulus, roots, logs (Examples 47 to 51).
5. Nature and integer roots (Examples 5 to 9).

JEE Advanced 2026 asked about integer roots with coefficients in A.P. (Example 9) and transformed roots of $x^2\pm x+1=0$ (Example 24); JEE Advanced 2025 used the odd degree fact (Example 55).

Quick recall

Practise these methods on real exam questions in the doMath Quadratic Equations chapter. Inequalities, the modulus and logarithms used above are developed in the Basic Algebra and Logarithms note, and the range of a rational function is used again in the Functions note.