Solve the inequality with square root
$$\sqrt{x+2}-\sqrt{x-1}\sqrt{2x-3} $$
for real values of $$x$$

Question

Solve the inequality with square root 
$$\sqrt{x+2}-\sqrt{x-1}>\sqrt{2x-3}      $$
for real values of $$x$$

Uniteasy Admin · Accepted Answer

Since $$\sqrt{2x-3}\geq 0 $$, squaring both sides of the inequality will not change the direction of the sign of the inequality.
$$x+2-2\sqrt{x+2}\sqrt{x-1}+x-1>2x-3 $$
Simplifying the squared inequality, we get
$$\sqrt{x+2}\sqrt{x-1}< 2$$
Since the LHS of square roots is greater than 0, we can square again without changing the sign of the inequality.
$$(x+2)(x-1) <4$$
Rearranging the inequality,
$$x^2+x-6<0$$
Solving the the quadratic gives
$$-3 < x<2$$n
In the meantime,
since the radicand of a square root is not less than 0, 
for $$\sqrt{x+2} $$ to be real, $$x+2\geq 0,$$
for $$\sqrt{x-1} $$ to be real , $$x-1\geq 0, $$
for $$\sqrt{2x-3} $$ to be real, $$2x-3\geq 0 $$
Therefore, the range of (1) is further narrowed to
$$\dfrac{3}{2}\leq x<2  $$

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