Question

solve $(x + 4)^2 - 3(x + 4) - 3 = 0$ using substitution.
$u = \boldsymbol{x + 4}$
select the solution(s) of the original equation.
$\frac{3+\sqrt{21}}{2}+4$
$\frac{3+\sqrt{21}}{2}-4$
$\frac{3-\sqrt{21}}{2}+4$
$\frac{3-\sqrt{21}}{2}-4$
retry

Explanation:

Step1: Substitute $u=x+4$

$u^2 - 3u - 3 = 0$

Step2: Apply quadratic formula

For $au^2+bu+c=0$, $u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=1, b=-3, c=-3$:
$u=\frac{3\pm\sqrt{(-3)^2-4(1)(-3)}}{2(1)}=\frac{3\pm\sqrt{9+12}}{2}=\frac{3\pm\sqrt{21}}{2}$

Step3: Substitute back $u=x+4$

Solve for $x$: $x = u - 4 = \frac{3\pm\sqrt{21}}{2}-4$

Answer:

B. $\frac{3+\sqrt{21}}{2}-4$
D. $\frac{3-\sqrt{21}}{2}-4$