Question

If $\alpha$ and $\beta$ are the roots of the equation $a{x}^2+bx+c=0$. The equation whose roots are as given below.
$\frac{1}{a\alpha +b},\frac{1}{a\beta +b}$ is ac$x^2$ - bx + 1 = 0. Her $a\alpha +b$ and $a\beta +b$ not equal to zero

• A. True
• B. False

Answer 1

The correct option is A True

$\Rightarrow$ $\alpha$ and $\beta$ are roots of the equation $a{x}^2+bx+c=0$

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}$ ----- ( 1 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}$ ------ ( 2 )

Now,

$\Rightarrow$ $\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}=\frac{a\beta +b+a\alpha +b}{(a\alpha +b)(a\beta +b)}$

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

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

$=\frac{a\times \frac{-b}{a}+2b}{a^2\times \frac{c}{a}+ab\times \frac{-b}{a}+b^2}$ [ Using ( 1 ) and ( 2 ) ]

$=\frac{b}{ac}$

$\therefore$ $\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}=\frac{b}{ac}$ -------- ( 3 )

$\Rightarrow$ $\frac{1}{a\alpha +b}.\frac{1}{a\beta +b}=\frac{1}{a^2\alpha \beta +ab\alpha +ab\beta +b^2}$

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

$=\frac{1}{a^2\times \frac{c}{a}+ab\times \frac{(-b)}{a}+b^2}$

$=\frac{1}{ac}$

$\Rightarrow$ $\frac{1}{a\alpha +b}.\frac{1}{a\beta +b}=\frac{1}{ac}$ ------ ( 4 )

New equation,

$\Rightarrow$ $x^2-(\frac{1}{a\alpha +b}+\frac{1}{a\beta +b})x+(\frac{1}{a\alpha +b}.\frac{1}{a\beta +b})=0$

$\Rightarrow$ $x^2-\frac{b}{ac}x+\frac{1}{ac}=0$

$\Rightarrow$ $ac{x}^2-bx+1=0$