Question

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$36x^2-12ax+\left(a^2-b^2\right)=0$

Answer 1

The given equation is $36x^2-12ax+\left(a^2-b^2\right)=0$.

Comparing it with $A{x}^2+Bx+C=0$, we get

A = 36, B = $-12a$ and C = $a^2-b^2$

∴ Discriminant, D = $B^2-4AC={\left(-12a\right)}^2-4\times 36\times \left(a^2-b^2\right)=144a^2-144a^2+144b^2=144b^2>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{144b^2}=12b$

$$\begin{gathered} \therefore \alpha =\frac{-B+\sqrt{D}}{2A}=\frac{-\left(-12a\right)+12b}{2\times 36}=\frac{12\left(a+b\right)}{72}=\frac{a+b}{6} \\ \beta =\frac{-B-\sqrt{D}}{2A}=\frac{-\left(-12a\right)-12b}{2\times 36}=\frac{12\left(a-b\right)}{72}=\frac{a-b}{6} \end{gathered}$$

Hence, $\frac{a+b}{6}$ and $\frac{a-b}{6}$ are the roots of the given equation.