Question

question 4
solve the radical equation, and check all proposed solutions.
\\(\sqrt{16x - 32} = x + 2\\)
\\(\circ\\) {6}
\\(\circ\\) {-5}
\\(\circ\\) {7}
\\(\circ\\) {-6}

Explanation:

Step1: Square both sides to eliminate the square root

To solve the radical equation \(\sqrt{16x - 32}=x + 2\), we first square both sides of the equation. Squaring a square root will eliminate the radical, so we get:
\((\sqrt{16x - 32})^2=(x + 2)^2\)
Simplifying both sides, the left side becomes \(16x - 32\) and the right side expands using the formula \((a + b)^2=a^2+2ab + b^2\) where \(a = x\) and \(b = 2\). So:
\(16x-32=x^{2}+4x + 4\)

Step2: Rearrange into standard quadratic form

Next, we rearrange the equation to form a standard quadratic equation \(ax^{2}+bx + c = 0\). Subtract \(16x\) and add \(32\) to both sides:
\(0=x^{2}+4x + 4-16x + 32\)
Simplify the right side:
\(x^{2}-12x + 36 = 0\)

Step3: Solve the quadratic equation

The quadratic equation \(x^{2}-12x + 36 = 0\) is a perfect square trinomial, which factors as \((x - 6)^2=0\).
Taking the square root of both sides, we get:
\(x-6 = 0\)
So, \(x = 6\)

Step4: Check the solution

Now we need to check if \(x = 6\) is a valid solution by substituting it back into the original equation.
Left side: \(\sqrt{16(6)-32}=\sqrt{96 - 32}=\sqrt{64}=8\)
Right side: \(6 + 2=8\)
Since both sides are equal, \(x = 6\) is a valid solution.

Answer:

\(\{6\}\)