What is sqrt(2d+1)+sqrt(d)<= 5 ?

The solution to sqrt(2d+1)+sqrt(d)<= 5 is 0<= d<= 4

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sqrt(2d+1)+sqrt(d)<= 5

2 d + 1 + d \leq 5

0 \leq d \leq 4

Interval Notation

[0, 4]

Solution steps

2 d + 1 + d \leq 5

Subtract

d

from both sides

2 d + 1 + d - d \leq 5 - d

Simplify

2 d + 1 \leq 5 - d

f (x) \leq g (x)

then

g (x) \geq 0

5 - d \geq 0 : d \leq 25

2 d + 1 \leq 5 - d : d \leq 4

Find singularity points

Find non-negative values for radicals:

d \geq 0

Combine the intervals

d \leq 25 and d \leq 4 and d \geq 0

Merge Overlapping Intervals

0 \leq d \leq 4

Verify Solutions:

0 \leq d \leq 4

True

The solution is

0 \leq d \leq 4

- 10 x - 10 = - 6 x - 2

5 \cdot \frac{2}{3}

x^{2} + 7 x + 10

43 = 64

16 = x - 13

What is sqrt(2d+1)+sqrt(d)<= 5 ?
The solution to sqrt(2d+1)+sqrt(d)<= 5 is 0<= d<= 4