Question

solve the radical equation $x - 7 = \sqrt{-4x + 28}$. which statement is true about the solutions to the radical equation?
there are no true solutions.
there are two true solutions.
there is one extraneous solution, with a value greater than 4.
there is one true solution, with a value greater than 6.

Explanation:

Step1: Square both sides to eliminate the square root

To solve the radical equation \( x - 7=\sqrt{-4x + 28} \), we first square both sides of the equation. Squaring the left side \((x - 7)\) gives \((x - 7)^2=x^{2}-14x + 49\), and squaring the right side \(\sqrt{-4x + 28}\) gives \(-4x + 28\). So we have the equation:
\(x^{2}-14x + 49=-4x + 28\)

Step2: Rearrange the equation to form a quadratic equation

Move all terms to one side of the equation to get a quadratic equation in standard form \(ax^{2}+bx + c = 0\). Subtract \(-4x + 28\) from both sides:
\(x^{2}-14x + 49+4x - 28 = 0\)
Simplify the left side:
\(x^{2}-10x + 21 = 0\)

Step3: Factor the quadratic equation

We factor the quadratic equation \(x^{2}-10x + 21 = 0\). We need two numbers that multiply to \(21\) and add up to \(-10\). The numbers are \(-3\) and \(-7\). So the factored form is:
\((x - 3)(x - 7)=0\)

Step4: Solve for x

Set each factor equal to zero and solve for \(x\):

• If \(x - 3=0\), then \(x = 3\)
• If \(x - 7=0\), then \(x = 7\)

Step5: Check for extraneous solutions

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

• For \(x = 3\):

Left side: \(3-7=-4\)
Right side: \(\sqrt{-4(3)+28}=\sqrt{-12 + 28}=\sqrt{16}=4\)
Since \(-4
eq4\), \(x = 3\) is an extraneous solution.

• For \(x = 7\):

Left side: \(7-7 = 0\)
Right side: \(\sqrt{-4(7)+28}=\sqrt{-28 + 28}=\sqrt{0}=0\)
Since \(0 = 0\), \(x = 7\) is a valid solution. And \(7>6\).

Answer:

There is one true solution, with a value greater than 6. (Corresponding to the option: There is one true solution, with a value greater than 6.)