Question

1. solve for all possible values of x. check for extraneous solutions.\\( x + 3 = \sqrt{3x + 13} \\)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \( x + 3=\sqrt{3x + 13} \), we first square both sides. Squaring the left side \((x + 3)\) gives \((x + 3)^2=x^{2}+6x + 9\), and squaring the right side \(\sqrt{3x + 13}\) gives \(3x + 13\). So we have the equation:
\[
x^{2}+6x + 9=3x + 13
\]

Step2: Rearrange into standard quadratic form

Subtract \(3x\) and \(13\) from both sides to get a quadratic equation in standard form \(ax^{2}+bx + c = 0\).
\[
x^{2}+6x-3x + 9 - 13=0
\]
Simplifying the like - terms, we have:
\[
x^{2}+3x - 4=0
\]

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(-4\) and add up to \(3\). The numbers are \(4\) and \(-1\). So we can factor the quadratic as:
\[
(x + 4)(x - 1)=0
\]

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:

• For \(x+4 = 0\), we get \(x=-4\).
• For \(x - 1=0\), we get \(x = 1\).

Step5: Check for extraneous solutions

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

Check \(x=-4\):

Left - hand side (LHS): \(x + 3=-4 + 3=-1\)
Right - hand side (RHS): \(\sqrt{3x + 13}=\sqrt{3\times(-4)+13}=\sqrt{-12 + 13}=\sqrt{1}=1\)
Since \(-1
eq1\), \(x=-4\) is an extraneous solution.

Check \(x = 1\):

Left - hand side (LHS): \(x + 3=1+3 = 4\)
Right - hand side (RHS): \(\sqrt{3x + 13}=\sqrt{3\times1+13}=\sqrt{3 + 13}=\sqrt{16}=4\)
Since \(4 = 4\), \(x = 1\) is a valid solution.

Answer:

The only solution is \(x = 1\) (and \(x=-4\) is extraneous).