What is sqrt(3-x)<sqrt(2x+1) ?

The solution to sqrt(3-x)<sqrt(2x+1) is 2/3 <x<= 3

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sqrt(3-x)<sqrt(2x+1)

3 - x < 2 x + 1

\frac{2}{3} < x \leq 3

Interval Notation

(\frac{2}{3}, 3]

Decimal Notation

0.66666 \dots < x \leq 3

Solution steps

3 - x < 2 x + 1

Square both sides

(3 - x)^{2} < (2 x + 1)^{2}

Simplify

3 - x < 2 x + 1

3 - x < 2 x + 1 : x > \frac{2}{3}

x > \frac{2}{3}

Find singularity points

Find non-negative values for radicals:

- \frac{1}{2} \leq x \leq 3

Combine the intervals

x > \frac{2}{3} and - \frac{1}{2} \leq x \leq 3

Merge Overlapping Intervals

\frac{2}{3} < x \leq 3

0 = x^{2} + 6 x + 5

\frac{7}{12} - \frac{3}{8}

\frac{3}{4} x - \frac{2}{3} = x + \frac{5}{6}

15 x^{5} - 18 x

4 (2 a + 2) - 17 < 15

What is sqrt(3-x)<sqrt(2x+1) ?
The solution to sqrt(3-x)<sqrt(2x+1) is 2/3 <x<= 3