What is 4sqrt(x+1)>= 12 ?

The solution to 4sqrt(x+1)>= 12 is x>= 8

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4sqrt(x+1)>= 12

4 x + 1 \geq 12

x \geq 8

Interval Notation

[8, \infty)

Solution steps

4 x + 1 \geq 12

Divide both sides by

4

\frac{4 x + 1}{4} \geq \frac{12}{4}

Simplify

x + 1 \geq 3

Square both sides

(x + 1)^{2} \geq 3^{2}

Simplify

x + 1 \geq 9

Move

1

to the right side

x \geq 8

Find singularity points

Find non-negative values for radicals:

x \geq - 1

Combine the intervals

x \geq 8 and x \geq - 1

Merge Overlapping Intervals

x \geq 8

\frac{5 2 + 3}{5 x + 3} < 1

\frac{1 0 ^{- 2}}{1 0 ^{- 3}}

\frac{13744}{100} x^{2} + \frac{54977}{100} x + 122500 = 0

218

(x + 4) (x + 5) = 0