Questions: Solve each equation. Remember to check for extraneous solutions. 2) sqrt(2x+10) = x+5 Identify your a, h, k. Then sketch the graph.

Solution Steps

Given the equation:

\[ \sqrt{2x+10} = x+5 \]

First, square both sides to eliminate the square root:

\[ (\sqrt{2x+10})^2 = (x+5)^2 \]

This simplifies to:

\[ 2x + 10 = x^2 + 10x + 25 \]

Rearrange the equation to bring all terms to one side:

\[ x^2 + 10x + 25 - 2x - 10 = 0 \]

Simplify:

\[ x^2 + 8x + 15 = 0 \]

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 8\), and \(c = 15\):

\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 15}}{2 \cdot 1} \]

Calculate the discriminant:

\[ b^2 - 4ac = 64 - 60 = 4 \]

Calculate the roots:

\[ x = \frac{-8 \pm \sqrt{4}}{2} \]

\[ x = \frac{-8 \pm 2}{2} \]

The solutions are:

\[ x = \frac{-8 + 2}{2} = -3 \]

\[ x = \frac{-8 - 2}{2} = -5 \]

Substitute \(x = -3\) back into the original equation:

\[ \sqrt{2(-3) + 10} = -3 + 5 \]

\[ \sqrt{4} = 2 \]

This is true, so \(x = -3\) is a valid solution.

Substitute \(x = -5\) back into the original equation:

\[ \sqrt{2(-5) + 10} = -5 + 5 \]

\[ \sqrt{0} = 0 \]

This is true, so \(x = -5\) is also a valid solution.

Final Answer