Question

Using quadratic formula solve for $x$:

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

Answer 1

Step 1: Comparing the given equation with the general form of the quadratic equation

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

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

Comparing the above two equations,

$$\begin{gathered} a=36 \\ b=-12a \\ c=a^2-b^2 \end{gathered}$$

Step 2: Substituting the values in the formula

The formula for finding the value of $x$ in a quadratic equation is

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

Upon substituting the above values we get,

$$\begin{gathered} \Rightarrow x=\frac{-(-12a)\pm \sqrt{{(-12a)}^2-4\times 36\times (a^2-b^2)}}{2\times 36} \\ \Rightarrow x=\frac{12a\pm \sqrt{144a^2-144a^2+144b^2}}{72} \\ \Rightarrow x=\frac{12a\pm \sqrt{144b^2}}{72} \\ \Rightarrow x=\frac{12a\pm 12b}{72} \end{gathered}$$

$$\begin{gathered} \Rightarrow x=\frac{12(a\pm b)}{72} \\ \Rightarrow x=\frac{a\pm b}{6} \end{gathered}$$

Therefore,

$x=\frac{a+b}{6}$ and $x=\frac{a-b}{6}$

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