What is (9-2x)^2<= (4x)^2 ?

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(9-2x)^2<= (4x)^2

(9 - 2 x)^{2} \leq (4 x)^{2}

x \leq - \frac{9}{2} or x \geq \frac{3}{2}

Interval Notation

(- \infty, - \frac{9}{2}] \cup [\frac{3}{2}, \infty)

Decimal Notation

x \leq - 4.5 or x \geq 1.5

Solution steps

(9 - 2 x)^{2} \leq (4 x)^{2}

Rewrite in standard form

- 4 x^{2} - 12 x + 27 \leq 0

Factor

- 4 x^{2} - 12 x + 27 : - (2 x - 3) (2 x + 9)

- (2 x - 3) (2 x + 9) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (2 x - 3) (2 x + 9)) (- 1) \geq 0 \cdot (- 1)

Simplify

(2 x - 3) (2 x + 9) \geq 0

Identify the intervals

x \leq - \frac{9}{2} or x \geq \frac{3}{2}

x (b + c) + 3 b + 3 c

lo g_{10} (2)

\frac{( 19 ) 13. 5 ^{2}}{32.8523}

2 x^{2} y^{3} - 4 x y^{4}

What is (9-2x)^2<= (4x)^2 ?
The solution to (9-2x)^2<= (4x)^2 is x<=-9/2 \lor x>= 3/2