Question

graph $y = 3sqrt3{(x + 1)^2}$ and $y = 20$ to check the solutions of $x = 7$ and $x = -9$.

Explanation:

Step1: Substitute x=7 into the first equation

Substitute $x=7$ into $y=5\sqrt{(x+1)^2}$:
$y=5\sqrt{(7+1)^2}=5\sqrt{8^2}=5\times8=40$

Step2: Compare with y=20

$40
eq20$, so $x=7$ is not a solution.

Step3: Substitute x=-9 into the first equation

Substitute $x=-9$ into $y=5\sqrt{(x+1)^2}$:
$y=5\sqrt{(-9+1)^2}=5\sqrt{(-8)^2}=5\times8=40$

Step4: Compare with y=20

$40
eq20$, so $x=-9$ is not a solution.

Step5: Analyze graph intersection

The graph of $y=5\sqrt{(x+1)^2}$ simplifies to $y=5|x+1|$, and $y=20$. Set $5|x+1|=20$, so $|x+1|=4$. This gives $x+1=4$ or $x+1=-4$, so $x=3$ or $x=-5$. These are the x-values where the graphs intersect.

Answer:

Neither $x=7$ nor $x=-9$ are solutions to $5\sqrt{(x+1)^2}=20$. The actual solutions (where the graphs intersect) are $x=3$ and $x=-5$.