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

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

Using ( 1 ) and ( 2 ),

$\Rightarrow$ ${\left(\frac{-b}{a}\right)}^2={\alpha }^2+{\beta }^2+2\times \frac{c}{a}$

$\Rightarrow$ $\frac{b^2}{a^2}={\alpha }^2+{\beta }^2+\frac{2c}{a}$

$\therefore$ ${\alpha }^2+{\beta }^2=\frac{b^2}{a^2}-\frac{2c}{a}$

$\therefore$ ${\alpha }^2+{\beta }^2=\frac{b^2-2ac}{a^2}$ ------ ( 3 )

Now,

$\Rightarrow$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{\frac{b^2-2ac}{a^2}}{\frac{c}{a}}$ [ Using ( 1 ) and ( 3 ) ]

$\Rightarrow$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{b^2-2ac}{ac}$ ----- ( 4 )

$\Rightarrow$ $\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=1$ ---- ( 5 )

New equation,

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

By Using ( 4 ) and ( 5 ),

$\Rightarrow$ $x^2-\left(\frac{b^2-4ac}{ac}\right)x+1=0$

$\Rightarrow$ $ac{x}^2-(b^2-4ac)x+1=0$

$\therefore$ We can see equation given in question is correct.