Question

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

Answer 1

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

Answer 2

um not totally sure

Answer 3

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

Answer 4

can you help please

Answer 5

What do you get if you multiply
\[(x^2-4)(x-3)\]?

Answer 6

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?

Answer 7

We know that:

\[= (\sqrt{x^2}x- \sqrt{x^2}3)^2 - (\sqrt{4}x-3 \sqrt{4})^2\]

Answer 8

no substitutions necessary. : )

Answer 9

\[0 = (x^2 - 3x)^2 - (2x - 6)^2\]

Answer 10

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

Answer 11

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

Answer 12

@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
\[(x-3)^2(x^2-4) =0\]