What is 4|x^2-1|+|x^2-4|>= 6 ?

The solution to 4|x^2-1|+|x^2-4|>= 6 is x = sqrt(2)

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4|x^2-1|+|x^2-4|>= 6

4 x^{2} - 1 + x^{2} - 4 \geq 6

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

Solution steps

4 x^{2} - 1 + x^{2} - 4 \geq 6

Find positive and negative intervals

x \leq - 2, - 2 < x \leq - 1, - 1 < x < 1, 1 \leq x < 2, x \geq 2

Solve the inequality for each interval

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

Merge Overlapping Intervals

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

- 3 ∣ 6 - x ∣ < - 15

∣ x - 1 ∣ + ∣ x - 2 ∣ + 2 ∣ x - 3 ∣ < 0

\frac{( x - 1 )}{( x - 2 )} \leq 3

\frac{( 3 - 4 x )}{5} \leq 4

34 + 21 x - x^{2} \leq - 1

What is 4|x^2-1|+|x^2-4|>= 6 ?
The solution to 4|x^2-1|+|x^2-4|>= 6 is x<=-sqrt(2)\lor-sqrt(2/5)<= x<= sqrt(2/5)\lor x>= sqrt(2)