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Quadratic Equations — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSQuadratic Equations1 Mark
MCQ
If $\alpha,\beta$ are the roots of the equation $ax^2+ bx + c = 0$, then $\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}$
  • A
    $\frac{\text{c}}{\text{ab}}$
  • B
    $\frac{\text{a}}{\text{bc}}$
  • ✓
    $\frac{\text{b}}{\text{ac}}$
  • D
    None of these.
✓

Answer

Correct option: C.
$\frac{\text{b}}{\text{ac}}$
  1. $\frac{\text{b}}{\text{ac}}$
Solution:
Given equation: $ax^2 + bx + c = 0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots $=\alpha+\beta=-\frac{\text{b}}{\text{a}}$
Product of the roots $=\alpha\beta=\frac{\text{c}}{\text{a}}$
$\therefore\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}=\frac{\text{a}\beta+\text{b}+\text{a}\alpha+\text{b}}{(\text{a}\alpha+\text{b})(\text{a}\beta+\text{b})}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}\alpha+\text{ab}\beta+\text{b}^2}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}(\alpha+\beta)+\text{b}^2}$
$=\frac{\text{a}\big(-\frac{\text{b}}{\text{a}}\big)+2\text{b}}{\text{a}^2\big(\frac{\text{c}}{\text{a}}\big)+\text{ab}\big(-\frac{\text{b}}{\text{a}}\big)+\text{b}^2}$
$=\frac{\text{b}}{\text{ac}}$

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