1 Answer

Sivanraj M · Answer 1 · 2021-03-11T15:00:35+0000

Answer:

$x=\pm 7\text{ or } x=\pm i√(2)$

Explanation:

We have the equation:

x^4-5x^2-14=0

Since this is in quadratic form, we can consider using u-substitution. Thus, we will let:

u=x^2

Then by substitution:

u^2-5x-14=0

Now we can solve normally. Factor:

(u-7)(u+2)=0

Zero Product Property:

$u=7\text{ or } u=-2$

Back-substitute:

$x^2=7\text{ and } x^2=-2$

Take the square root of both sides. Since we are taking an even-root, we will need to add plus-minus:

$x=\pm 7\text{ and }x=√(-2)$

Therefore, our solutions are:

$x=\pm7\text{ and } x=\pm i√(2)$

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