Question

Solve: $\frac{1}{\sqrt{4+3x-2x^2}}$

Answer 1

We have,

$\frac{1}{\sqrt{4+3x-2x^2}}$

Then,

$\frac{1}{\sqrt{-2x^2+3x+4}}$

$\Rightarrow \frac{1}{\sqrt{-2(x^2+\frac{3}{2}x+2)}}$

Make completing square, and we get,

$\frac{1}{\sqrt{-2}\sqrt{(x^2+\frac{3}{2}x+\frac{9}{16}+2-\frac{9}{16})}}$

$\Rightarrow \frac{1}{\sqrt{-2}\sqrt{{(x+\frac{3}{4})}^2+\frac{32-9}{16}}}$

$\Rightarrow \frac{1}{\sqrt{-2}\sqrt{{(x+\frac{3}{4})}^2+\frac{21}{16}}}$

Hence, this is the answer.