Find $\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$

Question

Find $$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$$

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My attempt :-

$$3(x-1)+2\sqrt{2x^2 - 7x - 4}$$ $$\implies 3x - 3 + 2\sqrt{2x^2 - 7x - 4} + 2x^2 - 2x^2 -7x + 7x - 1 +1 $$ $$\implies 2x^2 -7x - 4 + 2\sqrt{2x^2 - 7x - 4} + (- 2x^2 + 10x + 1) $$

let $y = \sqrt{2x^2 - 7x - 4}$

$$\therefore y^2 + 2y + (- 2x^2 + 10x + 1)$$

solving for $y$ :-

$$y = {-2\pm \sqrt{4 - 4\times(- 2x^2 + 10x + 1)} \over 2} $$ $$\implies y = {-1\pm \sqrt{2x^2 - 5x}} $$

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$$\therefore y^2 + 2y + (- 2x^2 + 10x + 1) = (y + 1 + \sqrt{2x^2 - 5x})(y + 1 - \sqrt{2x^2 - 5x})$$

$$\therefore \sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}} = \sqrt{(y + 1 + \sqrt{2x^2 - 5x})(y + 1 - \sqrt{2x^2 - 5x})\over 2}$$

Sadly this does not remove the square root.

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So the question is how can i factorize $3(x-1)+2\sqrt{2x^2 - 7x - 4}$ to remove that square root ?

just some hints are fine with me, thanks.

Answer 1

$$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}=$$ $$=\sqrt{\frac{3(x-1)+2\sqrt{2x^2 - 7x - 4}}2}=$$ $$=\sqrt\frac{\color{red}{3x-3}+2\sqrt{(2x+1)(x-4)}}2=$$ $$=\sqrt{\frac{\color{red}{2x+1}+2\sqrt{(2x+1)(x-4)}+\color{red}{x-4}}2}=$$ $$=\sqrt{\frac{(\sqrt{2x+1}+\sqrt{x-4})^2}2}=$$ $$=\frac{\sqrt{2x+1}+\sqrt{x-4}}{\sqrt2}$$

Answer 2

Hint: $2 x^2 - 7 x - 4 = (2x+1)(x-4)$, so you may guess that the desired result is of the form $a \sqrt{2x+1} + b \sqrt{x-4}$. Now square that, and see what $a$ and $b$ would give you $\dfrac{3}{2}(x-1) + \sqrt{2x^2-7x-4}$.