Question

What are the solutions of the equation x4 – 9x2 + 8 = 0? Use u substitution to solve.

Answer 1

Answer:

$(x ^ 2 -1)(x ^ 2 -8)\\\\x = 2\sqrt{2}\\\\x = - 2\sqrt{2}\\\\x = 1\\\\x = -1$

Step-by-step explanation:

We have the following equation $x^4 - 9x2 + 8 = 0$

To solve this problem, do:

$u = x ^ 2$

then it has:

$u ^ 2 - 9u + 8 = 0$

Then he has a second-degree equation.

To resolve factor.

You must find 2 numbers that when you add them get -9 and multiply them get 8.

These numbers are:

u = -1 and u = -8

Then it has:

$(u-1)(u-8) = 0$

$u = 1$

and

$u = 8$

Remember that $u = x ^ 2$

Then it has:

$(x ^ 2 -1)(x ^ 2 -8)\\\\x = 2\sqrt{2}\\\\x = - 2\sqrt{2}\\\\x = 1\\\\x = -1$

Answer 2

ANSWER

$x = \pm 2\sqrt{2} \: or \: {x}= \pm1$

EXPLANATION

Given;

${x}^{4} - 9 {x}^{2} +8 = 0$

We can rewrite as

$({x}^{2} )^{2} - 9 {x}^{2} +8 = 0$

Let

$u = {x}^{2}$

This implies that,

$u^{2} - 9u +8 = 0$

This implies that,

$u^{2}-u - 8u +8 = 0$

$u(u - 1) - 8(u - 1) = 0$

$(u - 1)(u - 8) = 0$

$u = 8 \: or \: u = 1$

But

$u = {x}^{2}$

This implies that,

${x}^{2} = 8 \: or \: {x}^{2} = 1$

$x = \pm \sqrt{8} \: or \: {x}= \pm \sqrt{1}$

$x = \pm 2\sqrt{2} \: or \: {x}= \pm1$