Question

solve the radical equation $7 - \sqrt{4x + 5} = x + 3$ (1 point)

$x = \square$

Explanation:

Step1: Isolate the radical term

Subtract \(x\) and \(3\) from both sides, or first isolate the square root. Let's move the radical to one side:
\(7 - x - 3=\sqrt{4x + 5}\)
Simplify the left side: \(4 - x=\sqrt{4x + 5}\)

Step2: Square both sides

To eliminate the square root, square both sides of the equation \((4 - x)^2 = (\sqrt{4x + 5})^2\)
Expand the left side: \(16 - 8x+x^2=4x + 5\)

Step3: Rearrange into quadratic equation

Bring all terms to one side: \(x^2-8x - 4x+16 - 5 = 0\)
Simplify: \(x^2-12x + 11 = 0\)

Step4: Solve the quadratic equation

Factor the quadratic: \((x - 1)(x - 11)=0\)
Set each factor equal to zero: \(x - 1 = 0\) or \(x - 11 = 0\)
So \(x = 1\) or \(x = 11\)

Step5: Check for extraneous solutions

• For \(x = 1\):

Left side of original equation: \(7-\sqrt{4(1)+5}=7-\sqrt{9}=7 - 3 = 4\)
Right side: \(1 + 3 = 4\). So \(x = 1\) is valid.

• For \(x = 11\):

Left side: \(7-\sqrt{4(11)+5}=7-\sqrt{49}=7 - 7 = 0\)
Right side: \(11 + 3 = 14\). \(0
eq14\), so \(x = 11\) is extraneous.

Answer:

\(x = 1\)