# How do you find the zeroes of f(x) = x^4= -7x^2 -144?

Jul 20, 2015

I found:
${x}_{1} = 4$
${x}_{2} = - 4$
${x}_{3} = 3 i$
${x}_{4} = - 3 i$

#### Explanation:

I start supposing that the second $=$ sign is not necessary, so:
$f \left(x\right) = {x}^{4} - 7 {x}^{2} - 144$
You can find the zeroes ($x$ values that makes your function equal to zero) by setting your function equal to zero and get:
${x}^{4} - 7 {x}^{2} - 144 = 0$

set ${x}^{2} = u$ so you get:
${u}^{2} - 7 u - 144 = 0$
Using the Quadratic Formula you get:
${u}_{1 , 2} = \frac{7 \pm \sqrt{49 - 4 \left(- 144\right)}}{2} = \frac{7 \pm 25}{2}$
so you get 2 solutions:
${u}_{1} = 16$
${u}_{2} = - 9$
But ${x}^{2} = u$ so $x = \pm \sqrt{u}$

You get 4 zeroes for your function (2 Real and 2 Immaginary):
${x}_{1} = \sqrt{16} = 4$
${x}_{2} = - \sqrt{16} = - 4$
${x}_{3} = \sqrt{- 9} = 3 i$
${x}_{4} = - \sqrt{- 9} = - 3 i$