## anonymous 4 years ago Find the two real-number solutions of x^4=1

1. JamesJ

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

2. anonymous

I don't understand how to do it

3. anonymous

+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.

4. JamesJ

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 $x = \pm 1 .$

5. anonymous

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

6. JamesJ

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.