• April 1st 2011, 09:25 PM
Ilsa
Find the range of values of $x$for which $(1+x)(6-x) < = -8$
• April 1st 2011, 11:05 PM
Prove It
Quadratic Inequalities are ALWAYS easiest to solve by completing the square...

$\displaystyle (1 + x)(6 - x) \leq -8$

$\displaystyle 6 - x + 6x - x^2 \leq -8$

$\displaystyle x^2 - 5x - 6 \geq 8$

$\displaystyle x^2 - 5x \geq 14$

$\displaystyle x^2 - 5x + \left(-\frac{5}{2}\right)^2 \geq 14 + \left(-\frac{5}{2}\right)^2$

$\displaystyle \left(x - \frac{5}{2}\right)^2 \geq \frac{56}{4} + \frac{25}{4}$

$\displaystyle \left(x - \frac{5}{2}\right)^2 \geq \frac{81}{4}$.

You should be able to go from here. Also, note that " $\displaystyle \sqrt{X^2} = |X|$" and " $\displaystyle |X| \geq a$" is equivalent to " $\displaystyle X \leq -a$ and $\displaystyle X \geq a$".