How is $\frac{1}{\sqrt{-2+3x-x²}} = \frac{2}{\sqrt{1-(2x-3)²}}$?

Question

I was trying to find the derivative of $arcsin (2x-3)$ when i noticed that my result differed from the solution given in my book. I went to check with wolfram alpha which tells me that my format is correct as well.

However I fail to see how these two functions are equal. I can't rework my result to the one given in my book and I fail to see why.

My result is the second one in the title.

Answer 1

$$\frac{2}{\sqrt{1-(2x-3)^2}} = \frac{2}{\sqrt{1 - 4x^2 + 12x - 9}} = \frac{2}{\sqrt{-8 + 12 x - 4x^2}} $$ If you divide by $2$ on the top and $\sqrt 4$ on the bottom: $$\frac{2}{\sqrt{-8 + 12 x - 4x^2}} = \frac{1}{\sqrt{-2 + 3 x - x^2}} $$

Answer 2

Hint. One may write $$ -2+3x-x^2=-\left[x^2-3x+2 \right]=-\left[\left(x-\frac32\right)^2-\frac94+2 \right]=-\frac14\left[\left(2x-3\right)^2-1 \right] $$ then one may apply the square root.

Answer 3

$$-2+3x-x^2 =\frac{-8+12x-4x^2}4$$ $$=\frac{-8-(4x^2-12x)}4$$ $$=\frac{-8+9-(4x^2-12x+9)}4$$ $$=\frac{1-(2x-3)^2}4$$ Now take square roots and reciprocals.