sovle for x. (x^2-1)^2-2(x^2-1)=3?

Question

Please show me the steps

Answer 1

(x^2-1)^2 - 2(x^2-1) = 3
lets expand out the first term and multiply the 2 into the 2nd term
(x^2-1)(x^2-1) - 2x^2 + 2 = 3
now, multiply out the x^2-1's
x^4-x^2-x^2+1-2x^2+2=3
combine terms and move the +3 to the other side (making it 0)
x^4-4x^2=0
now, we can see that x^2 is in both, so we'll pull that out
x^2(x^2-4) = 0
next, we know that x^2-4 is actually the binomial expansion of (x+2)(x-2), so we'll replace it with that.
x^2(x+2)(x-2)=0
and now, we have our answers of 0, -2, and 2.

Answer 2

3

Answer 3

(x² - 1)² - 2(x² - 1) = 3

First do all the indicated operations on the left side.

(x² - 1)² = x^4 - 2x² + 1.
-2(x² - 1) = -2x² + 2.

Now set the sum of these equal to 3.

x^4 - 2x² + 1 - 2x² + 2 = 3.

Combine like terms.

x^4 - 4x² + 3 = 3 ----> x^4 - 4x² = 0

We can factor out x².

x^4 - 4x² = x² (x² - 4) = 0

x² - 4 is also factorable.

x² - 4 = (x + 2)(x - 2)

Now set each factor above equal to 0.

x² = 0 ----> x = 0
x + 2 = 0 ----> x = -2
x - 2 = 0 ----> x = 2

Now substitute each value into the original equation to see if they make it true. If they do, you're done. I shall leave that as an exercise for you.