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anonymous

  • 4 years ago

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

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  1. JamesJ
    • 4 years ago
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    You can even guess the answers. What's at last one of them?

  2. anonymous
    • 4 years ago
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    I don't understand how to do it

  3. anonymous
    • 4 years ago
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    +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
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    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.

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