Question

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

Answer 1

`-3 - 2√(6)`

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 √(24) = \pm 2√(6)`

So
`x = -3\pm2√(6)`. Therefore, the smallest solution is
`x = -3-2√(6)`.

Answer 2

`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)= (-bб √(D) )/(2a) \\ x_1=(-6+ √(96) )/(2)= (-6+4 √(6) )/(2)= (2(2 √(6)-3 ))/(2)=2 √(6)-3 \\ x_2 =(-6- √(96) )/(2)= (-6-4 √(6) )/(2)= (2(-3-2 √(6)))/(2)=-3-2 √(6) \ \ \ smallest \ value`