What is (x+5)^2<= (x+4)^2+(x-3)^2 ?

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(x+5)^2<= (x+4)^2+(x-3)^2

(x + 5)^{2} \leq (x + 4)^{2} + (x - 3)^{2}

x \leq 0 or x \geq 8

Interval Notation

(- \infty, 0] \cup [8, \infty)

Solution steps

(x + 5)^{2} \leq (x + 4)^{2} + (x - 3)^{2}

Rewrite in standard form

- x^{2} + 8 x \leq 0

Factor

- x^{2} + 8 x : - x (x - 8)

- x (x - 8) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (x - 8)) (- 1) \geq 0 \cdot (- 1)

Simplify

x (x - 8) \geq 0

Identify the intervals

x \leq 0 or x \geq 8

x^{2} - 4 x - 25 > 20

2 x^{2} - 11 x = 0

2 (x + 5) - 1 = 2 x

\frac{6 ( x - 11 )}{5} = \frac{5 ( x + 11 )}{6}

\frac{3}{6} + \frac{10}{12}

What is (x+5)^2<= (x+4)^2+(x-3)^2 ?
The solution to (x+5)^2<= (x+4)^2+(x-3)^2 is x<= 0\lor x>= 8