Quadratic Equation #1

By factoring the first term, we see: $[4(x^2-2)]^2=x^2-2$ . Assume $p=x^2-2$ , then we can rewrite the equation as $(4p)^2=p$ .

Expanding, we get: $16p^2-p=0$ ,

And by factoring: $p(16p-1)=0$

Therefore, $\displaystyle p=0\ \text{or}\ {{1}\over{16}}$

Since $p=x^2-2$ , we can get the solutions to the equation by evaluating $x^2-2=0$ and $\displaystyle x^2-2={1\over 16}$ .

$x^2-2=0$

$x^2=2$

$x=±\sqrt{2}$

$\displaystyle x^2-2={1\over 16}$

$\displaystyle x^2={33\over 16}$

$\displaystyle x=±{\sqrt{33}\over 4}$

Now we just find the product of the positive solutions and add the product of the negative solutions to it.

$\displaystyle {{\sqrt{2}\times \sqrt{33}}\over 4}+{{-\sqrt{2}\times(-\sqrt{33})}\over {4}}$

$\displaystyle =2{\sqrt{66}\over{4}}={\sqrt{66}\over 2}\approx 4.1$