Question

Solve: $\mid x^2-3x+4\mid <12$

Answer 1

$\mid x^2-3x+4\mid <12$

$-12<x^2-3x+4<12$

(1) $0<x^2-3x+4+12$

$\Rightarrow x^2-3x+16>0$

$D\equiv 9-4\left(16\right)<0$ & $a>0$

$\Rightarrow x^2-3x+16>0\forall x\in R$

(2)$x^2-3x+4-12<0$

$\Rightarrow x^2-3x-8<0$

$\Rightarrow x-\frac{3\pm \sqrt{9+32}}{2}<0$

$\Rightarrow (x-\frac{3+\sqrt{4}1}{2})(x-\frac{3-\sqrt{4}1}{2})<0$

$\Rightarrow x\in (\frac{3-\sqrt{4}1}{2},\frac{3+\sqrt{4}1}{2})$