Question

solve the following radical equation. enter integers or reduced fractions as your answer. do not use decimals. if there is more than one solution, separate the solutions with a comma. if there is no solution, use dne. solve: \\(\sqrt{x + 6} = x + 4\\) solutions: \\(x = \square\\) extraneous solutions: \\(x = \square\\)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the radical equation \(\sqrt{x + 6}=x + 4\), we first square both sides of the equation. Squaring the left side \((\sqrt{x + 6})^2\) gives \(x + 6\), and squaring the right side \((x + 4)^2\) expands to \(x^2+8x + 16\) using the formula \((a + b)^2=a^2+2ab + b^2\) where \(a = x\) and \(b = 4\). So we get the equation:
\[x + 6=x^2+8x + 16\]

Step2: Rearrange into standard quadratic form

Subtract \(x\) and \(6\) from both sides to set the equation to zero:
\[x^2+8x + 16-x - 6 = 0\]
Simplify the left side:
\[x^2+7x + 10 = 0\]

Step3: Factor the quadratic equation

We need to factor \(x^2+7x + 10\). We look for two numbers that multiply to \(10\) and add up to \(7\). The numbers \(2\) and \(5\) work. So we can factor the quadratic as:
\[(x + 2)(x + 5)=0\]

Step4: Solve for x

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:

• If \(x+2=0\), then \(x=-2\).
• If \(x + 5=0\), then \(x=-5\).

Step5: Check for extraneous solutions

We need to check these solutions in the original equation \(\sqrt{x + 6}=x + 4\) because squaring both sides can introduce extraneous solutions.

• For \(x=-2\):

Left side: \(\sqrt{-2 + 6}=\sqrt{4}=2\)
Right side: \(-2 + 4 = 2\)
Since the left side equals the right side, \(x=-2\) is a valid solution.

• For \(x=-5\):

Left side: \(\sqrt{-5 + 6}=\sqrt{1}=1\)
Right side: \(-5 + 4=-1\)
Since \(1
eq - 1\), \(x=-5\) is an extraneous solution.

Answer:

Solutions: \(x=-2\)
Extraneous Solutions: \(x=-5\)