Solve this equation $(2x^2-3)^2=4(x-1)^2$?

Question

This is the way I solved:

What should I do next? Should I factorize or take a t that represents something?

Answer 1

$$\begin{align*} 0 &= (2x^2-3)^2 - 4(x-1)^2 = (2x^2-3)^2 - (2x-2)^2 \\ &= ((2x^2-3)-(2x-2))((2x^2-3)+(2x-2)) \\ &= (2x^2-2x-1)(2x^2+2x-5). \end{align*}$$ Now solve each individual quadratic; e.g., by completing the square or the quadratic formula.

Answer 2

You wouldn't have to factorize if you didn't expand squares before.

See this:

$(2x^2-3)^2=4(x-1)^2$

$(2x^2-3)^2-4(x-1)^2=0$

$\left[(2x^2-3)-2(x-1)\right]\left[(2x^2-3)+2(x-1)\right]=0$

$\left[2x^2-3-2x+2\right]\left[2x^2-3+2x-2\right]=0$

$(2x^2-2x-1)(2x^2+2x-5)=0$

$(x^2-x-1/2)(x^2+x-5/2)=0$

$(x^2-x+1/4-3/4)(x^2+x+1/4-11/4)=0$

$\left[(x-1/2)^2-3/4)\right]\left[(x+1/2)^2-11/4\right]=0$

so

$x = 1/2 \pm \sqrt3/2$ or $x=-1/2 \pm \sqrt{11}/2$.