Question

Using quadratic formula solve for $x$:

$9x^2-6a^{2}x+(a^4-b^4)=0$

Answer 1

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

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

The given equation is $9x^2-6a^{2}x+(a^4-b^4)=0$.

Comparing the above two equations,

$$\begin{gathered} a=9 \\ b=-6a^2 \\ c=a^4-b^4 \end{gathered}$$

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

The quadratic formula is

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

Substituting the values into the formula,

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

On simplifying,

$$\begin{gathered} \Rightarrow x=\frac{6a^2\pm \sqrt{36b^4}}{18} \\ \Rightarrow x=\frac{6a^2\pm 6b^2}{18} \\ \Rightarrow x=\frac{6(a^2\pm b^2)}{18} \\ \Rightarrow x=\frac{a^2\pm b^2}{3} \end{gathered}$$

The above equation can be written as,

$x=\frac{a^2+b^2}{3}$ and $x=\frac{a^2-b^2}{3}$.

Hence, we can conclude that the values for $x$ are $\frac{a^2+b^2}{3}$ and $\frac{a^2-b^2}{3}$.