Solve the equation $a^{2} x^{2} - 3 a b x + 2 b^{2} = 0$ by completing the square.

Open in App

Solution

There is nothing to prove if a = 0. For $a \neq 0$ , we have
$a^{2} x^{2} - 3 a b x + 2 b^{2} = 0$
$\to x^{2} - \frac{3 b}{a} x + \frac{2 b^{2}}{a^{2}} = 0$ $\to x^{2} - 2 (\frac{3 b}{2 a}) x = \frac{- 2 b^{2}}{a^{2}}$
$\to x^{2} - 2 (\frac{3 b}{2 a}) x + \frac{9 b^{2}}{4 a^{2}} - \frac{2 b^{2}}{a^{2}}$
$\to {(x - \frac{3 b}{2 a})}^{2} = \frac{9 b^{2} - 8 b^{2}}{4 a^{2}} \to {(x - \frac{3 b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}}$
$\to x - \frac{3 b}{2 a} = \pm \frac{b}{2 a} \to x = \frac{3 b \pm b}{2 a}$
Therefore, the solution set is ${\frac{b}{a}, \frac{2 b}{a}}$ .

Suggest Corrections

Similar questions

a^{2} x^{2} - 3 a b x + 2 b^{2} = 0

\frac{2 b}{a}, \frac{b}{a}

a^{2} x^{2} - 3 a b x + 2 b^{2} = 0

a^{2} x^{2} - 3 a b x + 2 b^{2} = 0, x = \frac{a}{b}, x = \frac{b}{a}

a^{2} x^{2} - 3 a b x + 2 b^{2} = 0

a^{2} x^{2} - 3 a b x + 2 b^{2}

x = \frac{a}{b} a n d x = \frac{b}{a}

Join BYJU'S Learning Program