What is sqrt(2x)+5<= sqrt(2x+35) ?

The solution to sqrt(2x)+5<= sqrt(2x+35) is No Solution for x\in\mathbb{R}

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sqrt(2x)+5<= sqrt(2x+35)

2 x + 5 \leq 2 x + 35

No Solution for x \in R

Solution steps

2 x + 5 \leq 2 x + 35

Move

2 x + 35

to the left side

2 x + 5 - 2 x + 35 \leq 0

Move

5

to the right side

2 x + 5 - 2 x + 35 \leq - 5

Subtract

5

from both sides

2 x + 5 - 2 x + 35 - 5 \leq - 5 - 5

Simplify

2 x - 2 x + 35 \leq - 10

Add

2 x + 35

to both sides

2 x - 2 x + 35 + 2 x + 35 \leq - 10 + 2 x + 35

Simplify

2 x \leq - 10 + 2 x + 35

f (x) \leq g (x)

then

g (x) \geq 0

- 10 + 2 x + 35 \geq 0 : x \geq \frac{65}{2}

2 x \leq - 10 + 2 x + 35 : x \leq \frac{169}{32}

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

x \geq \frac{65}{2} and x \leq \frac{169}{32} and x \geq 0

Merge Overlapping Intervals

No Solution for x \in R

7 z = 7

\frac{6 x + 7}{3} = \frac{9 x + 3}{7}

(x + 6) (x - 7) \leq 0

c^{2} = 6^{2} + 8^{2}

3 (2)

What is sqrt(2x)+5<= sqrt(2x+35) ?
The solution to sqrt(2x)+5<= sqrt(2x+35) is No Solution for x\in\mathbb{R}