Question

find the solution to the equation, or explain why there is no solution. remember to check for extraneous solutions. \\(\sqrt{3x + 16} = x + 2\\)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{3x + 16}=x + 2\), we first square both sides. Squaring the left side gives us \(3x + 16\), and squaring the right side \((x + 2)^2\) expands to \(x^2+4x + 4\) using the formula \((a + b)^2=a^2+2ab + b^2\) where \(a = x\) and \(b = 2\). So we get the equation:

$$3x+16=x^{2}+4x + 4$$

Step2: Rearrange into standard quadratic form

Subtract \(3x\) and \(16\) from both sides to set the equation to zero:

$$x^{2}+4x + 4-3x - 16=0$$

Simplify the like terms: \(4x-3x=x\) and \(4 - 16=-12\), so the equation becomes:

$$x^{2}+x-12 = 0$$

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(- 12\) and add up to \(1\). The numbers are \(4\) and \(-3\) since \(4\times(-3)=-12\) and \(4+(-3) = 1\). So we can factor the quadratic as:

$$(x + 4)(x-3)=0$$

Step4: Solve for x using zero - product property

If \((x + 4)(x - 3)=0\), then either \(x+4 = 0\) or \(x - 3=0\). Solving \(x+4 = 0\) gives \(x=-4\), and solving \(x - 3=0\) gives \(x = 3\).

Step5: Check for extraneous solutions

• For \(x=-4\): Substitute into the original equation \(\sqrt{3(-4)+16}\) and \(-4 + 2\). The left side is \(\sqrt{-12 + 16}=\sqrt{4}=2\), and the right side is \(-2\). Since \(2

eq-2\), \(x = - 4\) is an extraneous solution.

• For \(x = 3\): Substitute into the original equation. The left side is \(\sqrt{3(3)+16}=\sqrt{9 + 16}=\sqrt{25}=5\), and the right side is \(3+2 = 5\). Since \(5 = 5\), \(x = 3\) is a valid solution.

Answer:

\(x = 3\)