What is 5x^2+11x>= 3x^2+40 ?

The solution to 5x^2+11x>= 3x^2+40 is x = 5/2

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5x^2+11x>= 3x^2+40

5 x^{2} + 11 x \geq 3 x^{2} + 40

x \leq - 8 or x \geq \frac{5}{2}

Interval Notation

(- \infty, - 8] \cup [\frac{5}{2}, \infty)

Decimal Notation

x \leq - 8 or x \geq 2.5

Solution steps

5 x^{2} + 11 x \geq 3 x^{2} + 40

Rewrite in standard form

2 x^{2} + 11 x - 40 \geq 0

Factor

2 x^{2} + 11 x - 40 : (2 x - 5) (x + 8)

(2 x - 5) (x + 8) \geq 0

Identify the intervals

x \leq - 8 or x \geq \frac{5}{2}

- (0)^{2}

4 x^{2} = - 24

3 x + 4 = 25

∣ 2 x + 4 ∣ = 2

(x - 3) (x + 1)^{2} (x - 8) \geq 0

What is 5x^2+11x>= 3x^2+40 ?
The solution to 5x^2+11x>= 3x^2+40 is x<=-8\lor x>= 5/2