#### Question

Solve the inequality with square root

\sqrt{x+2}-\sqrt{x-1}>\sqrt{2x-3}

for real values of x

Collected in the board: Inequality

Steven Zheng posted 3 months ago

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

-3 < x<2
(1)

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

Steven Zheng posted 3 months ago

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