# How do you solve the inequality x^2<=abs(4x-3) and write your answer in interval notation?

Oct 1, 2017

${x}^{2} \le \left\mid 4 x - 3 \right\mid$

so, we have either ${x}^{2} \le 4 x - 3$ or ${x}^{2} \le - \left(4 x - 3\right)$

i.e. ${x}^{2} - 4 x + 3 \le 0$ or ${x}^{2} + 4 x - 3 \le 0$

solving ${x}^{2} - 4 x + 3 = 0$, we $x = 1 \mathmr{and} x = 3$

Hence, $1 \le x \le 3$

Solving ${x}^{2} + 4 x - 3 = 0$ using the quadratic formula gives:

$x = - 2 \pm \sqrt{7}$

Hence, $- 2 - \sqrt{7} \le x \le - 2 + \sqrt{7}$

Using interval notation we have:

$\left[- 2 - \sqrt{7} , - 2 + \sqrt{7}\right]$ OR $\left[1 , 3\right]$

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