What is sqrt(x+7)>= sqrt(2x-1) ?

The solution to sqrt(x+7)>= sqrt(2x-1) is 1/2 <= x<= 8

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sqrt(x+7)>= sqrt(2x-1)

x + 7 \geq 2 x - 1

\frac{1}{2} \leq x \leq 8

Interval Notation

[\frac{1}{2}, 8]

Decimal Notation

0.5 \leq x \leq 8

Solution steps

x + 7 \geq 2 x - 1

Square both sides

(x + 7)^{2} \geq (2 x - 1)^{2}

Simplify

x + 7 \geq 2 x - 1

x + 7 \geq 2 x - 1 : x \leq 8

x \leq 8

Find singularity points

Find non-negative values for radicals:

x \geq \frac{1}{2}

Combine the intervals

x \leq 8 and x \geq \frac{1}{2}

Merge Overlapping Intervals

\frac{1}{2} \leq x \leq 8

0 = x^{2} + x - 2

6 x^{2} - 5 x + 1 = 0

(3^{2})^{- \frac{3}{2}}

x^{2} - 4 x + 4 = 2 x - 4

2 x^{2} + 8 x - 960 = 0

What is sqrt(x+7)>= sqrt(2x-1) ?
The solution to sqrt(x+7)>= sqrt(2x-1) is 1/2 <= x<= 8