Question

which of the following is an extraneous solution of \\(\sqrt{4x + 41} = x + 5\\)?

• (x = -8)
• (x = 2)
• (x = 8)
• (x = -2)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{4x + 41}=x + 5\), we first square both sides. Squaring the left side gives us \(4x + 41\), and squaring the right side gives us \((x + 5)^2=x^{2}+10x + 25\). So our equation becomes:
\[4x + 41=x^{2}+10x + 25\]

Step2: Rearrange into a quadratic equation

Subtract \(4x\) and \(41\) from both sides to set the equation to zero:
\[x^{2}+10x + 25-4x - 41 = 0\]
Simplify the left side:
\[x^{2}+6x - 16 = 0\]

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(- 16\) and add to \(6\). The numbers are \(8\) and \(-2\). So we can factor the quadratic as:
\[(x + 8)(x - 2)=0\]

Step4: Solve for \(x\)

Set each factor equal to zero:

• If \(x + 8=0\), then \(x=-8\)
• If \(x - 2=0\), then \(x = 2\)

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

Check \(x=-8\):

Left side: \(\sqrt{4(-8)+41}=\sqrt{-32 + 41}=\sqrt{9}=3\)
Right side: \(-8 + 5=-3\)
Since \(3
eq - 3\), \(x =-8\) is an extraneous solution.

Check \(x = 2\):

Left side: \(\sqrt{4(2)+41}=\sqrt{8 + 41}=\sqrt{49}=7\)
Right side: \(2+5 = 7\)
Since \(7 = 7\), \(x = 2\) is a valid solution.

Answer:

\(x=-8\)