Question

solve the equation. check for extraneous solutions.
\\(sqrt{x + 7} = x + 1\\)

\\(\circ\\) no solution

\\(\circ\\) 2

\\(\circ\\) -3

\\(\circ\\) -3, 2

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{x + 7}=x + 1\), we first square both sides. Squaring the left side gives us \(x + 7\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)), and squaring the right side gives us \((x + 1)^2\). So we have the equation:
\[x + 7=(x + 1)^2\]

Step2: Expand the right - hand side

We know that \((a + b)^2=a^2+2ab + b^2\). For \((x + 1)^2\), \(a = x\) and \(b = 1\), so \((x + 1)^2=x^2+2x + 1\). Our equation becomes:
\[x + 7=x^2+2x + 1\]

Step3: Rearrange the equation to form a quadratic equation

Subtract \(x\) and \(7\) from both sides to get:
\[x^2+2x + 1-x - 7=0\]
Simplify the left - hand side:
\[x^2+(2x - x)+(1 - 7)=0\]
\[x^2+x - 6=0\]

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(-6\) and add up to \(1\). The numbers are \(3\) and \(-2\). So we can factor \(x^2+x - 6\) as:
\((x + 3)(x - 2)=0\)

Step5: Solve for \(x\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So:
If \(x+3 = 0\), then \(x=-3\); if \(x - 2=0\), then \(x = 2\)

Step6: Check for extraneous solutions

• Check \(x=-3\):

Substitute \(x=-3\) into the original equation \(\sqrt{x + 7}=x + 1\). The left - hand side is \(\sqrt{-3 + 7}=\sqrt{4}=2\), and the right - hand side is \(-3 + 1=-2\). Since \(2
eq - 2\), \(x=-3\) is an extraneous solution.

• Check \(x = 2\):

Substitute \(x = 2\) into the original equation. The left - hand side is \(\sqrt{2+7}=\sqrt{9}=3\), and the right - hand side is \(2 + 1=3\). Since \(3 = 3\), \(x = 2\) is a valid solution.

Answer:

2