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anonymous

  • 4 years ago

How would you solve x^4=2x^2-1? Precalculus prerequisites

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  1. anonymous
    • 4 years ago
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    Start by factoring it.

  2. anonymous
    • 4 years ago
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    I believe so, it says solve the equation algebraically.. however, I feel braindead right now, it isn't clicking. I tried factoring, nothing goes into either.. does it?

  3. anonymous
    • 4 years ago
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    What about 1?

  4. anonymous
    • 4 years ago
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    As in if you are using the method of what goes into the constant that multiplys to get the constant in front of the x^4 term and also adds to get the middle term. So more or less -1 and -1...

  5. anonymous
    • 4 years ago
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    I'm beyond confused.. that doesn't sound familiar at all.. english please? haha

  6. anonymous
    • 4 years ago
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    Ok I'll just do whatever else does on here and maybe you'll see it from the answer.

  7. anonymous
    • 4 years ago
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    *left

  8. anonymous
    • 4 years ago
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    \[(x^2-1)(x^2-1)=0\]

  9. anonymous
    • 4 years ago
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    that's like precalc inception, I honestly would never think of that..

  10. anonymous
    • 4 years ago
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    There are different ways to factor, I've always begun with the method I tried to describe. If you look, the two negative ones multiply to get the constant of 1 in front of the x^4 and add to get the -2 in the middle.

  11. anonymous
    • 4 years ago
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    wait.. so it'd be like (x^2 - 1) (x^2 - 1) as you said, then.. how would you add the -2 if that's already a coefficient?

  12. anonymous
    • 4 years ago
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    that's where it doesn't make sense, because I always remembered when factoring you can't take out any coefficients unless your simplifying?

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