JIPMAT 2024 (QA) - If and are the roots of the equation a x^2+b x+c=0, then value of1/a +b+1/a +b is : | PYQs + Solutions

JIPMAT 2024

Algebra

Quadratic Equations

Hard

If $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$ , then value of
$\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}$ is :

\frac{a}{b c}

\frac{\mathrm{b}}{\mathrm{ac}}

\frac{c}{a b}

\frac{1}{a b c}

Correct Option: 2

1) Since

\alpha

and

\beta

are roots of

ax^2 + bx + c = 0

\newline

a\alpha^2 + b\alpha + c = 0

...(1)

\newline

a\beta^2 + b\beta + c = 0

...(2)

2) We need to find

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

3) From (1), we can write:

\newline

c = -a\alpha^2 - b\alpha

\newline

c = -a\beta^2 - b\beta

4) Let's find LCM:

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

\frac{(a\beta+b) + (a\alpha+b)}{(a\alpha+b)(a\beta+b)}

\frac{a(\alpha+\beta) + 2b}{(a\alpha+b)(a\beta+b)}

5) For quadratic equation

ax^2 + bx + c = 0

\newline

\alpha + \beta = -\frac{b}{a}

6) Substituting:

\newline

\frac{a(-\frac{b}{a}) + 2b}{(a\alpha+b)(a\beta+b)}

\frac{-b + 2b}{(a\alpha+b)(a\beta+b)}

\frac{b}{(a\alpha+b)(a\beta+b)}

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

Put

\alpha\beta=\frac{c}{a}

and

\alpha+\beta=\frac{-b}{a}

in the above equation, and we'll get

\frac{b}{ac}

\newline

Therefore,

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