What is [(2d+1)^{(1/2)}]+[(d)^{(1/2)}]<= 5 ?

The solution to [(2d+1)^{(1/2)}]+[(d)^{(1/2)}]<= 5 is 0<= d<= 4

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[(2d+1)^{(1/2)}]+[(d)^{(1/2)}]<= 5

[(2 d + 1)^{(\frac{1}{2})}] + [(d)^{(\frac{1}{2})}] \leq 5

0 \leq d \leq 4

Interval Notation

[0, 4]

Solution steps

[(2 d + 1)^{\frac{1}{2}}] + [d^{\frac{1}{2}}] \leq 5

Simplify

[(2 d + 1)^{\frac{1}{2}}] + [d^{\frac{1}{2}}] \leq 5 : [2 d + 1] + [d] \leq 5

[2 d + 1] + [d] \leq 5

2 d + 1 + d \leq 5

Move

d

to the right side

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

7 x - 5 = 23

4 = 4 (x + 10)

\frac{5 + e}{2} = - 10

\frac{4}{36}

4 (\frac{x - 2 x ^{2}}{3} + 2) = 2.5

What is [(2d+1)^{(1/2)}]+[(d)^{(1/2)}]<= 5 ?
The solution to [(2d+1)^{(1/2)}]+[(d)^{(1/2)}]<= 5 is 0<= d<= 4