What is 15x^4>= 4x^2+35 ?

The solution to 15x^4>= 4x^2+35 is x = sqrt(5/3)

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15x^4>= 4x^2+35

15 x^{4} \geq 4 x^{2} + 35

x \leq - \frac{5}{3} or x \geq \frac{5}{3}

Interval Notation

(- \infty, - \frac{5}{3}] \cup [\frac{5}{3}, \infty)

Decimal Notation

x \leq - 1.29099 \dots or x \geq 1.29099 \dots

Solution steps

15 x^{4} \geq 4 x^{2} + 35

Rewrite in standard form

15 x^{4} - 35 - 4 x^{2} \geq 0

Factor

15 x^{4} - 35 - 4 x^{2} : (5 x^{2} + 7) (3 x + 5) (3 x - 5)

(5 x^{2} + 7) (3 x + 5) (3 x - 5) \geq 0

Identify the intervals

x \leq - \frac{5}{3} or x \geq \frac{5}{3}

16 + \frac{1}{3}

\frac{270 π}{180}

- 11 x^{3} y (- 11 y^{3})

10 - 4 ∣ 2 x - 1 ∣ = 6

8 x - 6

What is 15x^4>= 4x^2+35 ?
The solution to 15x^4>= 4x^2+35 is x<=-sqrt(5/3)\lor x>= sqrt(5/3)