Question

(25 Points!!) :Find the smallest value of $x$ such that $x^2+6x + 9 = 24$.

Answer 1

$x^2+6x + 9 = 24 \\ x^2+6x + 9 - 24=0 \\ x^2+6x - 15 = 0 \\ D=b^2-4ac=6^2-4*1*(-15)=36+60=96 \\ x_{1,2}= \frac{-bб \sqrt{D} }{2a} \\ x_1=\frac{-6+ \sqrt{96} }{2}= \frac{-6+4 \sqrt{6} }{2}= \frac{2(2 \sqrt{6}-3 )}{2}=2 \sqrt{6}-3 \\ x_2 =\frac{-6- \sqrt{96} }{2}= \frac{-6-4 \sqrt{6} }{2}= \frac{2(-3-2 \sqrt{6})}{2}=-3-2 \sqrt{6} \ \ \ smallest \ value$

Answer 2

Answer:

$-3 - 2\sqrt{6}$

Step-by-step explanation:

The left side is the square of a binomial:

                                                 $(x + 3)^{2} = 24$

Taking the square root of both sides gives

                                                 $x + 3 = \pm \sqrt{24} = \pm 2\sqrt{6}$

So $x = -3\pm2\sqrt{6}$. Therefore, the smallest solution is $x = -3-2\sqrt{6}$.