Question

Question in pic! Please show work!

Answer 1

No, and there are a few ways to check his solution.

1. If $x=-13$, we get $x^2-6x-7=240\neq0$. So -13 can't be a solution, and the same goes for 19.

2. The quadratic can be factorized pretty easily: $x^2-6x-7=(x-7)(x+1)$, which means the solutions should be $x=7$ and $x=-1$.

3. Check Josh's reasoning. The mistake occurs between the 4th and 5th/7th lines, where Josh wrote

$(x-3)^2=16\implies x-3=16$

$(x-3)^2=16\implies x-3=-16$

This is not true. He was supposed to take the square root of 16 first:

$(x-3)^2=16\implies x-3=\sqrt{16}=4\implies x=7$

$(x-3)^2=16\implies x-3=-\sqrt{16}=-4\implies x=-1$

Answer 2

Answer:

Josh's solution is NOT correct.

Step-by-step explanation:

Josh's solution is NOT correct.

x^2 - 6x - 7 = 0

As you know:

-7 = (-7) * (+1) and -6 = (-7) + (+1) = -6

So factor x^2 - 6x - 7 = 0, you have

(x - 7)(x + 1) = 0

Set each factor = 0 to solve the equation

x - 7 = 0; x = 7

x + 1 = 0; x = -1

Solutions: x = -1 and x = 7

So Josh's solution is NOT correct.