Theory of EquationsmediumFree

Theory of Equations: Roots Equation Equal

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

Question
If α\alpha and β\beta are the roots of the equation ax2+bx+c=0ax^2 + bx + c = 0, then 1aα+b+1aβ+b\dfrac{1}{a\alpha + b} + \dfrac{1}{a\beta + b} is equal to
Abac\dfrac{b}{ac}correct
B−bac\dfrac{-b}{ac}
Cacb\dfrac{ac}{b}
D−acb\dfrac{-ac}{b}
Solution
Step 1: From Vieta's: α+β=−b/a\alpha + \beta = -b/a and αβ=c/a\alpha\beta = c/a. Step 2:
1aα+b+1aβ+b=(aα+b)+(aβ+b)(aα+b)(aβ+b)=a(α+β)+2ba2αβ+ab(α+β)+b2\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=ba(-b/a)+2b = -b+2b = b. Denominator: a2⋅ca+ab(−ba)+b2=ac−b2+b2=aca^2 \cdot \dfrac{c}{a} + ab\left(-\dfrac{b}{a}\right)+b^2 = ac - b^2 + b^2 = ac.
1aα+b+1aβ+b=bac\frac{1}{a\alpha+b}+\frac{1}{a\beta+b} = \frac{b}{ac}
 Correct answer: (1)
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