What is x^2+4x+4<= 7 ?

The solution to x^2+4x+4<= 7 is -sqrt(7)-2<= x<= sqrt(7)-2

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x^2+4x+4<= 7

x^{2} + 4 x + 4 \leq 7

- 7 - 2 \leq x \leq 7 - 2

Interval Notation

[- 7 - 2, 7 - 2]

Decimal Notation

- 4.64575 \dots \leq x \leq 0.64575 \dots

Solution steps

x^{2} + 4 x + 4 \leq 7

Factor

x^{2} + 4 x + 4 : (x + 2)^{2}

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

For

u^{n} \leq a

, if

n

is even then

- 7 \leq x + 2 \leq 7

a \leq u \leq b

then

a \leq u and u \leq b

- 7 \leq x + 2 and x + 2 \leq 7

- 7 \leq x + 2 : x \geq - 7 - 2

x + 2 \leq 7 : x \leq 7 - 2

Combine the intervals

x \geq - 7 - 2 and x \leq 7 - 2

Merge Overlapping Intervals

- 7 - 2 \leq x \leq 7 - 2

\frac{8 + 4 i}{2 i}

x^{4} + 2 x^{3} - 12 x^{2} - 40 x - 32

3 (x + 2) + 4 = 5 (x - 1) + x

5 (x + 4) = 35

\frac{5 x + 1}{1 + 4 x} \geq 0

What is x^2+4x+4<= 7 ?
The solution to x^2+4x+4<= 7 is -sqrt(7)-2<= x<= sqrt(7)-2