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farmergirl411

  • 3 years ago

Solve for x x^2(x-3)^2-4(x-3)^2=0

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  1. whpalmer4
    • 3 years ago
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    Can you rearrange that so it is a product? If you can, then set each component of the product = 0 and solve for x.

  2. farmergirl411
    • 3 years ago
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    um not totally sure

  3. farmergirl411
    • 3 years ago
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    can you solve it another way I am totally lost with this problem

  4. farmergirl411
    • 3 years ago
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    can you help please

  5. whpalmer4
    • 3 years ago
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    What do you get if you multiply \[(x^2-4)(x-3)\]?

  6. whpalmer4
    • 3 years ago
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    Let's make a substitution to make it easier to see what is going on. Instead of \[x^2(x-3)^2 - 4(x-3)^2 = 0\]We'll let u = (x-3) and write\[x^2u^2-4u^2 = 0\]Can you factor that at all?

  7. ambius
    • 3 years ago
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    We know that: \[= (\sqrt{x^2}x- \sqrt{x^2}3)^2 - (\sqrt{4}x-3 \sqrt{4})^2\]

  8. ambius
    • 3 years ago
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    no substitutions necessary. : )

  9. ambius
    • 3 years ago
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    \[0 = (x^2 - 3x)^2 - (2x - 6)^2\]

  10. ambius
    • 3 years ago
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    \[(x^2-3x)^2 = (2x-6)^2\] then root both sides, simplify, and solve for x.

  11. valentinekid
    • 3 years ago
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    @whpalmer4 had the right idea. Factor (x-3)^2 out: \[(x^2-4)(x-3)^2=0\]

  12. whpalmer4
    • 3 years ago
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    @ambius no substitution was *necessary*, but people often have trouble spotting that they can factor out a binomial rather than a single term, until they've done it a few times. Spotting that one can factor out a single term is something that many do more easily. It's easy to see that \[x^2u^2 - 4u^2 =0\]can be factored to \[u^2(x^2-4)=0\]and then when you plug \(u = (x-3)\) back in, you've got \[(x-3)^2(x^2-4) = 0\]Yes, someone more experienced at factoring of this sort will notice directly that \[x^2(x-3)^2 - 4(x-3)^2 = 0\]factors to \[(x-3)^2(x^2-4) =0\]

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