anonymous
  • anonymous
Solve for x x^2(x-3)^2-4(x-3)^2=0
Calculus1
jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
Can you rearrange that so it is a product? If you can, then set each component of the product = 0 and solve for x.
anonymous
  • anonymous
um not totally sure
anonymous
  • anonymous
can you solve it another way I am totally lost with this problem

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anonymous
  • anonymous
can you help please
whpalmer4
  • whpalmer4
What do you get if you multiply \[(x^2-4)(x-3)\]?
whpalmer4
  • whpalmer4
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?
anonymous
  • anonymous
We know that: \[= (\sqrt{x^2}x- \sqrt{x^2}3)^2 - (\sqrt{4}x-3 \sqrt{4})^2\]
anonymous
  • anonymous
no substitutions necessary. : )
anonymous
  • anonymous
\[0 = (x^2 - 3x)^2 - (2x - 6)^2\]
anonymous
  • anonymous
\[(x^2-3x)^2 = (2x-6)^2\] then root both sides, simplify, and solve for x.
anonymous
  • anonymous
@whpalmer4 had the right idea. Factor (x-3)^2 out: \[(x^2-4)(x-3)^2=0\]
whpalmer4
  • whpalmer4
@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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