Question

Solve $1\le \frac{3x^2-7x+8}{x^2+1}\le 2$

Answer 1

$1\le \frac{3x^2-7x+8}{x^2+1}\le 2$

$\frac{3x^2-7x+8-2(x^2+1)}{x^2+1}\le 0$

$(x^2+1>0)$

$3x^2-7x+8-2x^2-2\le 0$

$x^2-7x+6\le 0$

$x^2-6x-x+6\le 0$

$(x-1)(x-6)\le 0$

$x\epsilon (1,6)$

$\left(ii\right)1\le \frac{3x^2-7x+8}{x^2+1}$

$\frac{3x^2-7x+8-(x^2+1)}{x^2+1}\ge 0$

$3x^2-7x+8-x^2-x\ge 0$

$2x^2-7x+7\ge 0$

$D=b^2-4ac=49-(4\times 2\times 7)<0$

$hencex\epsilon allrealnumbers$

$from\left(i\right)\&\left(ii\right)$

$x\epsilon (1,6)$