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.
$\alpha +\frac{1}{\beta },\beta +\frac{1}{\alpha }$ is $ac{x}^2+b(a+c)x+(a+c{)}^2=0$

• 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,

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

$=(\alpha +\beta )+\left(\frac{\alpha +\beta }{\alpha \beta }\right)$

$=\frac{-b}{a}+\frac{\frac{-b}{a}}{\frac{c}{a}}$ [ By using ( 1 ) and ( 2 ) ]

$=\frac{-b}{a}-\frac{b}{c}$

$=\frac{-bc-ba}{ac}$

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

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

$=\frac{c}{a}+2+\frac{1}{\frac{c}{a}}$

$=\frac{c}{a}+2+\frac{a}{c}$

$=\frac{a^2+2ac+c^2}{ac}$

$\therefore$ $(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })=\frac{a^2+2ac+c^2}{ac}$ ----- ( 4 )

Now, new equation,

$\Rightarrow$ $x^2-(\alpha +\frac{1}{\beta }+\beta +\frac{1}{\alpha })x+\left[(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })\right]=0$

By using ( 3 ) and ( 4 ),

$\Rightarrow$ $x^2+\frac{b(a+c)}{ac}x+\frac{a^2+2ac+c^2}{ac}$

$\Rightarrow$ $ac{x}^2+b(a+c)x+(a+c{)}^2=0$