What is |x-1-x^2|<= |x^2-3x+4| ?

The solution to |x-1-x^2|<= |x^2-3x+4| is x<= 3/2

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|x-1-x^2|<= |x^2-3x+4|

x - 1 - x^{2} \leq x^{2} - 3 x + 4

x \leq \frac{3}{2}

Interval Notation

(- \infty, \frac{3}{2}]

Decimal Notation

x \leq 1.5

Solution steps

x - 1 - x^{2} \leq x^{2} - 3 x + 4

Find positive and negative intervals

- \infty < x < \infty

Solve the inequality for each interval

x \leq \frac{3}{2}

x^{2} - 5 ∣ x ∣ + 6 < 0

\frac{( 3 x )}{( x ^{2} - 4 )} \leq 1

∣ 5 x ∣ < ∣ 10 - 5 x ∣

x^{2} - 3 x + ∣ x + 1 ∣ \geq 4

\frac{( x - 4 )}{( x + 2 )} \leq \frac{( x - 2 )}{( x - 1 )}

What is |x-1-x^2|<= |x^2-3x+4| ?
The solution to |x-1-x^2|<= |x^2-3x+4| is x<= 3/2