Question

Using the quadratic formula, solve for $x$:

$9x^2-6ax+a^2-b^2=0$

Answer 1

Step 1: Comparing the given question with the standard form of the quadratic equation

The standard form of the quadratic equation is $a{x}^2+bx+c=0$.

The given equation is $9x^2-6ax+a^2-b^2=0$.

On comparing the two equations,

$$\begin{gathered} a=9 \\ b=-6a \\ c=a^2-b^2 \end{gathered}$$

Step 2: Substituting the values into the quadratic formula for finding $x$

The quadratic formula is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting values,

$$\begin{gathered} \Rightarrow x=\frac{-(-6a)\pm \sqrt{{(-6a)}^2-4\times 9\times (a^2-b^2)}}{2\times 9} \\ \Rightarrow x=\frac{6a\pm \sqrt{36a^2-36(a^2-b^2)}}{18} \\ \Rightarrow x=\frac{6a\pm \sqrt{36a^2-36a^2+36b^2}}{18} \end{gathered}$$

On simplifying,

$$\begin{gathered} \Rightarrow x=\frac{6a\pm \sqrt{36b^2}}{18} \\ \Rightarrow x=\frac{6a\pm 6b}{18} \\ \Rightarrow x=\frac{6(a\pm b)}{18} \\ \Rightarrow x=\frac{a\pm b}{3} \end{gathered}$$

The above equation can be written as,

$x=\frac{a+b}{3}$ and $x=\frac{a-b}{3}$.

Hence, the roots of the given equation are $\frac{a+b}{3}$ and $\frac{a-b}{3}$.