What is sqrt(d+3)+sqrt(d+7)>4 ?

The solution to sqrt(d+3)+sqrt(d+7)>4 is d>-3/4

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sqrt(d+3)+sqrt(d+7)>4

d + 3 + d + 7 > 4

d > - \frac{3}{4}

Interval Notation

(- \frac{3}{4}, \infty)

Decimal Notation

d > - 0.75

Solution steps

d + 3 + d + 7 > 4

Subtract

d + 7

from both sides

d + 3 + d + 7 - d + 7 > 4 - d + 7

Simplify

d + 3 > 4 - d + 7

f (x) > g (x)

then

g (x) < 0 or g (x) \geq 0

For

4 - d + 7 < 0 : d > 9

For

4 - d + 7 \geq 0 : - \frac{3}{4} < d \leq 9

Find singularity points

Find non-negative values for radicals:

d \geq - 3

Combine the intervals

(- \frac{3}{4} < d \leq 9 or d > 9) and d \geq - 3

Merge Overlapping Intervals

d > - \frac{3}{4}

Verify Solutions:

d > - \frac{3}{4}

True

The solution is

d > - \frac{3}{4}

9 p^{2} + 6 p - 8

6 x (x - 1) = 25 - 11 x

x^{2} = - 4

x^{2} + 50 = 0

\frac{1}{2} (- 2) - 2