# In the following, determine whether the given values are solutions of the given equation or not: $a^2x^2\ â€“\ 3abx\ +\ 2b^2\ =\ 0,\ x\ =\ \frac{a}{b},\ x\ =\ \frac{b}{a}$

Given:

The given equation is $a^2x^2\ –\ 3abx\ +\ 2b^2\ =\ 0$.

To do:

We have to determine whether $x=\frac{a}{b}, x=\frac{b}{a}$ are solutions of the given equation.

Solution:

If the given values are the solutions of the given equation then they should satisfy the given equation.

Therefore,

For $x=\frac{a}{b}$,

LHS$=a^2x^2 – 3abx + 2b^2$.

$=a^2(\frac{a}{b})^2-3ab(\frac{a}{b})+2b^2$

$=a^2(\frac{a^2}{b^2})-3a^2+2b^2$

$=\frac{a^4}{b^2}-3a^2+2b^2$

$≠$RHS

Hence, $x=\frac{a}{b}$ is not a solution of the given equation.

For $x=\frac{b}{a}$,

LHS$=a^2x^2 – 3abx + 2b^2$

$=a^2(\frac{b}{a})^2-3ab(\frac{b}{a})+2b^2$

$=a^2(\frac{b^2}{a^2})-3b^2+2b^2$

$=b^2-3b^2+2b^2$

$=0$

RHS$=0$

LHS$=$RHS

Hence, $x=\frac{b}{a}$ is a solution of the given equation.â€Š

Tutorialspoint

Simply Easy Learning

Advertisements