What is 2sqrt(x+1)<= 7 ?

The solution to 2sqrt(x+1)<= 7 is -1<= x<= 45/4

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2sqrt(x+1)<= 7

2 x + 1 \leq 7

- 1 \leq x \leq \frac{45}{4}

Interval Notation

[- 1, \frac{45}{4}]

Decimal Notation

- 1 \leq x \leq 11.25

Solution steps

2 x + 1 \leq 7

Divide both sides by

2

\frac{2 x + 1}{2} \leq \frac{7}{2}

Simplify

x + 1 \leq \frac{7}{2}

Square both sides

(x + 1)^{2} \leq (\frac{7}{2})^{2}

Simplify

x + 1 \leq (\frac{7}{2})^{2}

Expand

(\frac{7}{2})^{2} : \frac{49}{4}

x + 1 \leq \frac{49}{4}

Move

1

to the right side

x \leq \frac{45}{4}

Find singularity points

Find non-negative values for radicals:

x \geq - 1

Combine the intervals

x \leq \frac{45}{4} and x \geq - 1

Merge Overlapping Intervals

- 1 \leq x \leq \frac{45}{4}

- 7 x + 12 = 0

x^{2} + 3 = 5 x + 3

6 - 4 (- 9 v + 5) > - 2 v + 2 + 3 v

\frac{x}{6} = \frac{x}{7} - 10

- 3 \geq 2 x + 3 > 7