Question

Solve the inequality with square root

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

for real values of x

Collected in the board: Inequality

Steven Zheng posted 2 weeks ago


Answer

First for the square roots to be real values,

3-x\geq 0 and x+1\geq 0

The x is confined in the range of

-1\leq x\leq 3

Now let’s introduce an intermediate variable

Let t=\sqrt{x+1}, (0\leq t\leq 2)

Then, x=t^2-1

The original inequality can be converted to an inequality in terms of t

\sqrt{4-t^2}>\dfrac{1}{2}+t

Square both sides of the inequality

4-t^2>\dfrac{1}{4}+t+t^2

2t^2+t-\dfrac{15}{4}<0

Considering the domain of t,solve the quadratic inequality

0 < t<\dfrac{\sqrt{31}-1 }{4}

0 < t^2 < (\dfrac{\sqrt{31}-1 }{4} )^2

-1 < t^2-1 < (\dfrac{\sqrt{31}-1 }{4} )^2-1

-1 < x < (\dfrac{\sqrt{31}-1 }{4} )^2-1

-1 < x <1-\dfrac{\sqrt{31} }{8}

Steven Zheng posted 2 weeks ago

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