Theory of EquationsmediumFree

Theory of Equations: Roots Equation Equal

JEE Maths question with a full step-by-step solution.

Question

\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}

is equal to

\dfrac{b}{ac}

correct

\dfrac{-b}{ac}

\dfrac{ac}{b}

\dfrac{-ac}{b}

Solution

Step 1: From Vieta's:

\alpha + \beta = -b/a

and

\alpha\beta = c/a

. Step 2:

\frac{1}{a\alpha+b} + \frac{1}{a\beta+b} = \frac{(a\alpha+b)+(a\beta+b)}{(a\alpha+b)(a\beta+b)} = \frac{a(\alpha+\beta)+2b}{a^2\alpha\beta + ab(\alpha+\beta)+b^2}

Step 3: Numerator:

a(-b/a)+2b = -b+2b = b

. Denominator:

a^2 \cdot \dfrac{c}{a} + ab\left(-\dfrac{b}{a}\right)+b^2 = ac - b^2 + b^2 = ac

\frac{1}{a\alpha+b}+\frac{1}{a\beta+b} = \frac{b}{ac}

Correct answer: (1)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Mathsquestions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.