Set of all the values of x satisfying the inequality x^ 2 - 7x + 6...

x^{2} - 7x + 6 \ge 0 ⇒ x \in \left(- \infty,1\right] \cup \left[6, \infty\right)<br />Case I:x < - 2$<br />LHS is non negative, RHS is negative<br />$⇒$~the inequality is always true</p><p>$⇒ x < - 2 .$<br />Case II:$x \ge - 2$<br />$x^{2} - 7x + 6 > x^{2} + 4x + 4$<br />$- 11x > - 2 ⇒ x < \dfrac{2}{11}$<br />$⇒ x \in \left[-$2,$\dfrac{2}{11}\right)
Taking the union of the two cases, we get,~
$x \in \left(- \infty, \dfrac{2}{11}\right)</p>$