What is |2x^2-5x+1|>= 2 ?

The solution to |2x^2-5x+1|>= 2 is x = (sqrt(33)+5)/4

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|2x^2-5x+1|>= 2

2 x^{2} - 5 x + 1 \geq 2

x \leq \frac{- 33 + 5}{4} or 1 \leq x \leq \frac{3}{2} or x \geq \frac{33 + 5}{4}

Interval Notation

(- \infty, \frac{- 33 + 5}{4}] \cup [1, \frac{3}{2}] \cup [\frac{33 + 5}{4}, \infty)

Decimal Notation

x \leq - 0.18614 \dots or 1 \leq x \leq 1.5 or x \geq 2.68614 \dots

Solution steps

2 x^{2} - 5 x + 1 \geq 2

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

2 x^{2} - 5 x + 1 \leq - 2 or 2 x^{2} - 5 x + 1 \geq 2

2 x^{2} - 5 x + 1 \leq - 2 or 2 x^{2} - 5 x + 1 \geq 2

1 \leq x \leq \frac{3}{2} or (x \leq \frac{- 33 + 5}{4} or x \geq \frac{33 + 5}{4})

Merge Overlapping Intervals

x \leq \frac{- 33 + 5}{4} or 1 \leq x \leq \frac{3}{2} or x \geq \frac{33 + 5}{4}

∣ x ∣ + ∣ 2 x - 8 ∣ \leq ∣ x + 8 ∣

\frac{( 3 x + 1 )}{( x - 3 )} > 6

∣ z ∣ \leq 7

∣ 2 x + 5 ∣ > 0.5

- 4 x + 14 + 10 ∣ x - 2 ∣ < 12

What is |2x^2-5x+1|>= 2 ?
The solution to |2x^2-5x+1|>= 2 is x<= (-sqrt(33)+5)/4 \lor 1<= x<= 3/2 \lor x>= (sqrt(33)+5)/4