Question

Solve for all possible values of x.
square root of, 6, x, plus, 40, end square root, equals, x, plus, 8
6x+40
​
=
x+8
Answer
Attempt 1 out of 2

Answer 1

To solve the equation

\[
\sqrt{6x + 40} = x + 8,
\]

we will first square both sides to eliminate the square root.

\[
(\sqrt{6x + 40})^2 = (x + 8)^2.
\]

This simplifies to:

\[
6x + 40 = (x + 8)(x + 8).
\]

Now expand the right side:

\[
6x + 40 = x^2 + 16x + 64.
\]

Next, we will rearrange the equation to set it to zero:

\[
0 = x^2 + 16x + 64 - 6x - 40.
\]

Combine like terms:

\[
0 = x^2 + 10x + 24.
\]

Now we have a quadratic equation:

\[
x^2 + 10x + 24 = 0.
\]

We can factor this equation:

\[
(x + 4)(x + 6) = 0.
\]

Setting each factor equal to zero gives us the possible solutions:

1. \(x + 4 = 0 \implies x = -4\)
2. \(x + 6 = 0 \implies x = -6\)

Now we have two possible solutions: \(x = -4\) and \(x = -6\).

However, we need to check each solution in the original equation to make sure they are valid (since squaring can introduce extraneous solutions).

1. **Check \(x = -4\)**:

\[
\sqrt{6(-4) + 40} = -4 + 8 \implies \sqrt{-24 + 40} = 4 \implies \sqrt{16} = 4,
\]
which is true.

2. **Check \(x = -6\)**:

\[
\sqrt{6(-6) + 40} = -6 + 8 \implies \sqrt{-36 + 40} = 2 \implies \sqrt{4} = 2,
\]
which is also true.

Both solutions are valid. Thus, the final solutions are:

\[
\boxed{-4 \text{ and } -6}.
\]