Asked by d
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
square root of, 6, x, plus, 40, end square root, equals, x, plus, 8
6x+40
=
x+8
Answer
Attempt 1 out of 2
Answers
Answered by
GPT-4o mini
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:
\[
\boxed{-4 \text{ and } -6}.
\]
\[
\sqrt{6x + 40} = x + 8,
\]
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:
\[
\boxed{-4 \text{ and } -6}.
\]
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