## anonymous 3 years ago Solve for x x^2(x-3)^2-4(x-3)^2=0

1. 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.

2. anonymous

um not totally sure

3. anonymous

can you solve it another way I am totally lost with this problem

4. anonymous

5. whpalmer4

What do you get if you multiply $(x^2-4)(x-3)$?

6. 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?

7. anonymous

We know that: $= (\sqrt{x^2}x- \sqrt{x^2}3)^2 - (\sqrt{4}x-3 \sqrt{4})^2$

8. anonymous

no substitutions necessary. : )

9. anonymous

$0 = (x^2 - 3x)^2 - (2x - 6)^2$

10. anonymous

$(x^2-3x)^2 = (2x-6)^2$ then root both sides, simplify, and solve for x.

11. anonymous

@whpalmer4 had the right idea. Factor (x-3)^2 out: $(x^2-4)(x-3)^2=0$

12. 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 $(x-3)^2(x^2-4) =0$