Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
radical equations and extraneous roots
what is the solution of \\(\sqrt{x^2 + 49} = x + 5\\)?
\\(x = \frac{12}{5}\\)
\\(x = -\frac{12}{5}\\)
\\(x = -6\\) or \\(x = -3\\)
no solution

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{x^2 + 49}=x + 5\), we first square both sides of the equation. Squaring the left side gives us \(x^2+49\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)), and squaring the right side gives us \((x + 5)^2\). Using the formula \((a + b)^2=a^2+2ab + b^2\) with \(a=x\) and \(b = 5\), we have \((x + 5)^2=x^2+10x + 25\). So the equation becomes:
\[x^2+49=x^2+10x + 25\]

Step2: Simplify the equation

Subtract \(x^2\) from both sides of the equation. The \(x^2\) terms will cancel out:
\[x^2 - x^2+49=x^2 - x^2+10x + 25\]
\[49=10x + 25\]

Step3: Solve for \(x\)

Subtract 25 from both sides:
\[49-25=10x+25 - 25\]
\[24 = 10x\]
Then divide both sides by 10:
\[x=\frac{24}{10}=\frac{12}{5}\]

Step4: Check for extraneous solutions

We need to check if \(x = \frac{12}{5}\) satisfies the original equation. Substitute \(x=\frac{12}{5}\) into the left side: \(\sqrt{(\frac{12}{5})^2+49}=\sqrt{\frac{144}{25}+49}=\sqrt{\frac{144 + 1225}{25}}=\sqrt{\frac{1369}{25}}=\frac{37}{5}\). Substitute into the right side: \(\frac{12}{5}+5=\frac{12 + 25}{5}=\frac{37}{5}\). Since both sides are equal, \(x=\frac{12}{5}\) is a valid solution.

Answer:

\(x=\frac{12}{5}\) (corresponding to the option \(x=\frac{12}{5}\))