Theory of Equations29 questions

Theory of Equations — JEE Maths Practice Questions & Solutions

29 questions on Theory of Equations with full step-by-step solutions, including past-year (PYQ) problems. Free to practice.

expert

If roots of the equation

a(x-\ell)(x-m) + n = 0

are

\alpha, \beta

, then roots of the equation

a\!\left(\dfrac{x}{x-k} - \alpha\right)\!\left(\dfrac{x}{x-k} - \beta\right) - n = 0

are

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hard

a \in \mathbb{R}

and the equation

(a-2)(x-[x])^2 + 2(x-[x]) + a^2 = 0

(where

[x]

denotes GIF) has no integral solution and has exactly one solution in

(2, 3)

, then

a

\alpha

be a root of

ax^2 + bx + c = 0

and

\beta

be a root of

-ax^2 + bx + c = 0

, where

a, b, c

are real and

a \neq 0

. Then the equation

\dfrac{a}{2}x^2 + bx + c = 0

has a root

\gamma

such that

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medium

If one of the roots of

ax^2 + ax + a + 1 = 0

is less than

1

and the other is greater than

1

, then the complete set of values of

a

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medium

a, b, c, d

are non-zero real numbers such that

c

and

d

are the roots of

x^2 + ax + b = 0

and

a

and

b

are the roots of

x^2 + cx + d = 0

, then the absolute value of

a + 2b + 3c + 4d

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medium

\alpha, \beta, \gamma

be the roots of the equation

x^3 + ax + a = 0

(where

a \in \mathbb{R}

a \neq 0

) satisfying

\dfrac{\alpha^2}{\beta} + \dfrac{\beta^2}{\gamma} + \dfrac{\gamma^2}{\alpha} = -8

, then

a

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medium

\alpha + \dfrac{1}{\alpha}

and

\beta + \dfrac{1}{\beta}

are the zeroes of

f(x) = x^2 - 5x - a

where

\alpha, \beta \in (0, \infty)

, then the set of values of

a

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medium

\alpha

and

\beta

are the roots of the equation

x^2 - x + 11 = 0

, then the value of

3\alpha^3 - 3\alpha^2 + 2\beta^3 - 2\beta^2 + 11\alpha

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medium

\alpha

and

\beta

are the roots of the equation

x^2 - 2x - 5 = 0

, then the equation whose roots are

\dfrac{\alpha}{\alpha-1}

and

\dfrac{\beta}{\beta-1}

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medium

If the quadratic equations

3x^2 + ax + 1 = 0

and

2x^2 + bx + 1 = 0

have a common root, then the value of

|2a^2 - 5ab + 3b^2|

is, where

2a \neq 3b

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medium

(\alpha, \beta)

(\beta, \gamma)

(\gamma, \alpha)

are respectively the roots of

x^2 - 2px + 2 = 0

;

x^2 - 2qx + 3 = 0

;

x^2 - 2rx + 6 = 0

where

\alpha, \beta, \gamma

are all positive, then the value of

p + q + r

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medium

A value of

b

for which the equations

x^2 + bx - 1 = 0

and

x^2 + x + b = 0

have one root in common is

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medium

Let

\alpha, \beta

be the roots of the equation

(x-a)(x-b) = c

c \neq 0

. Then the roots of the equation

(x-\alpha)(x-\beta) + c = 0

a

for which

(a-1)x^2 - (a+1)x + (a-1) \geq 0

holds true for all

x \geq 2

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medium

a, b, c, d

are distinct positive real numbers such that

a

and

b

are the roots of

x^2 - 10cx - 11d = 0

and

c

and

d

are the roots of

x^2 - 10ax - 11b = 0

, then the value of

a + b + c + d

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medium

The number of integral values of

a

so that the graph of

y = 16x^2 + 8(a+5)x - 7a - 5

is always above the

x

-axis is

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medium

\alpha

and

\beta

are the roots of the equation

ax^2 + bx + c = 0

, then

\dfrac{1}{a\alpha + b} + \dfrac{1}{a\beta + b}

\alpha, \beta, \gamma, \delta

be the roots (real or non-real) of

x^4 - 3x + 1 = 0

. The value of

\alpha^3 + \beta^3 + \gamma^3 + \delta^3

is equal to

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medium

\tan^2\!\dfrac{3\pi}{8}

is a root of the equation

2x^2 - 3ax + 4b = 0

, where

a, b \in \mathbb{Q}

, then the value of

a + 4b

P(x) = x^2 - (2-p)x + p - 2

. If

P(x)

assumes both positive and negative values

\forall x \in \mathbb{R}

, then the range of

p

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medium

\alpha, \beta, \gamma

are roots of the equation

111x^3 - 11x - 1 = 0

, then

(\alpha\beta)^{-2} + (\beta\gamma)^{-2} + (\gamma\alpha)^{-2}

x^3 + 2x^2 - 4x + 5 = 0

has roots

\alpha, \beta, \gamma

, then the value of

\dfrac{(\alpha^3+5)(\beta^3+5)(\gamma^3+5)}{13\alpha\beta\gamma}

is equal to

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medium

c^2 = 4d

and the two equations

x^2 - ax + b = 0

and

x^2 - cx + d = 0

have one common root, then the value of

2(b+d)

\alpha

and

\beta

for which the quadratic equation

x^2 + 2x + 2 + e^{2\alpha} - \cos\beta = 0

has a real solution is

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easy

If the vertex of the quadratic expression

y = 2x^2 - 4x + 6

(m, n)

, then the quadratic equation whose vertex is

\left(2m, \dfrac{n}{2}\right)

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easy

\alpha, \beta

are the roots of

x^2 - p(x+1) - c = 0

c \neq 1

, then the value of

(\alpha+1)(\beta+1)

\alpha

and

\beta

be the roots of the equation

x^2 - (p-2)x - (p-1) = 0

p \in \mathbb{R}

. If

\alpha^2 + \beta^2

is least, then

p

equals

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easy

\alpha, \beta, \gamma

are roots of the equation

x^3 - 2x^2 - 1 = 0

and

T_n = \alpha^n + \beta^n + \gamma^n

, then the value of

\dfrac{T_{11} - T_8}{T_{10}}

is equal to

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easy

If the quadratic polynomial

P(x) = 4x^2 + (k+8)x + 9

k \in \mathbb{I}

, is positive for every real

x

, then the greatest absolute value of

k

View solution →

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