Question

Find the two real-number solutions of x^4=1

Answer 1

You can even guess the answers. What's at last one of them?

Answer 2

I don't understand how to do it

Answer 3

+1 and -1. This is because an even number exponent always has a positive answer. and you can do 1^1/4 and still get 1.

Answer 4

Right. Taking the square root of both sides once, you have
\[ \sqrt{x^4} = x^2 = \pm 1 \]

Now \( x^2 \geq 0 \) so it must be
\[ x^2 = 1\]
Take the square root again and we have
\[ x = \pm 1 .\]

Answer 5

ok i think i get it. the next problem is x^4=16/81 so do i just multiply 16/81 and 1/4?

Answer 6

No. Take the square root once you and you have
\[ \sqrt{x^4} = (x^4)^{1/2} = x^2 \]
and
\[ \pm \sqrt{ \frac{16}{81}} = \frac{\sqrt{16}}{\sqrt{81}} = \frac{4}{9} \]
Hence
\[ x^4 = \frac{16}{81} \implies x^2 = \pm \frac{4}{9} \]

Now, just like your last question, \( x^2 \geq 0 \) for all real numbers. Hence
\[ x^2 = \frac{4}{9} \]

Now take the square root again.