Question

$a^2{x}^2-3abx+2b^2$= 0 then $x=\frac{a}{b}andx=\frac{b}{a}$ are the roots of the equation.

• A. True
• B. False
• C. Ambiguos
• D. Can't say

Answer 1

Answer: (B)

False

The quadratic equation is
$f\left(x\right)=a^2{x}^2-3abx+2b^2=0$ (i)
Given,
$x=\frac{a}{b},x=\frac{b}{a}$
Putting $x=\frac{a}{b}$ in equation (i)
$f\left(\frac{a}{b}\right)=a^2(\frac{a}{b}{)}^2-3ab(\frac{a}{b})+2b^2$
$=\frac{a^4}{b^2}+2b^2-3a^2\ne 0$
Putting $x=\frac{b}{a}$ in equation (i) we get,
$f\left(\frac{b}{a}\right)=a^2(\frac{b}{a}{)}^2-3ab(\frac{b}{a})+2b^2$
$=b^2-3b^2+2b^2=0$
Thus,$x=\frac{b}{a}$ satisfies the quadratic equation $f\left(x\right)=a^2{x}^2-3abx+2b^2=0$and
$x=\frac{a}{b}$ is not a solution of $f\left(x\right)=a^2{x}^2-3abx+2b^2=0$