X^2(x+1)^2-14x(x+1)+24=0

[Trigarius]

Please help me to solve for x . Many thanks

 

[MarkFL]

To make things simpler, try letting \(\displaystyle u=x(x+1)\) and rewrite the equation in terms of \(\displaystyle u\). What do you get?

 

[MarkFL]

No, I was suggesting the above substitution so that the OP would have a quadratic solely in \(\displaystyle u\)[.

 

[stapel]

Please help me to solve for x . Many thanks

You don't include any exercise within your actual post so I will guess that your question is in the subject line. (Subject lines commonly do not relate to post contents, so please correct me if I'm wrong.) Also, I will guess that, contrary to standard notation, you actually mean "X" and "x" to mean the same thing.

I'm assuming that the question is as follows:

\(\displaystyle \mbox{Solve by factoring: }\, x^2(x\, +\, 1)^2\, -\, 14x(x\, +\, 1)\, +\, 24\, =\, 0\)

If so, then the hint provided earlier is spot-on. Using that substitution reduces the quadratic equation to:

. . . . .\(\displaystyle u^2\, -\, 14u\, +\, 24\, =\, 0\)

...since \(\displaystyle x^2(x\, +\, 1)^2\, =\, \left[x(x\, +\, 1)\right]^2\)

Can you factor the simplified quadratic?