Question

Using the quadratic formula, solve for $x$:

$16x^2-8a^{2}x+(a^4-b^4)=0$

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 $16x^2-8a^{2}x+(a^4-b^4)=0$.

By comparing the above two equations,

$$\begin{gathered} a=16 \\ b=-8a^2 \\ c=a^4-b^4 \end{gathered}$$

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

The quadratic equation is

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

After substituting the values, the equation becomes

$$\begin{gathered} \Rightarrow x=\frac{-\left(-8a^2\right)\pm \sqrt{{(-8a^2)}^2-4\times 16\times (a^4-b^4)}}{2\times 16} \\ \Rightarrow x=\frac{8a^2\pm \sqrt{64a^4-64a^4+64b^4}}{32} \\ \Rightarrow x=\frac{8a^2\pm \sqrt{64b^4}}{32} \\ \Rightarrow x=\frac{8a^2\pm 8b^2}{32} \end{gathered}$$

Taking $8$ as common in both numerator and denominator

$$\begin{gathered} \Rightarrow x=\frac{8(a^2\pm b^2)}{32} \\ \Rightarrow x=\frac{a^2\pm b^2}{4} \end{gathered}$$

Therefore,

$x=\frac{a^2+b^2}{4}$ and $x=\frac{a^2-b^2}{4}$

Hence, the values of $x$ for the given equation are $\frac{a^2+b^2}{4}$ and $\frac{a^2-b^2}{4}$.