Question

solve the radical equation. check your solution(s). write
o solution\ if there is no solution.
$2 + \sqrt{4x - 3} = x$
$x = \square$

Explanation:

Step1: Isolate the radical

Subtract 2 from both sides of the equation \(2 + \sqrt{4x - 3}=x\) to get \(\sqrt{4x - 3}=x - 2\).

Step2: Square both sides

Square both sides to eliminate the square root: \((\sqrt{4x - 3})^2=(x - 2)^2\), which simplifies to \(4x - 3=x^2 - 4x + 4\).

Step3: Rearrange into quadratic equation

Rearrange the equation to standard quadratic form: \(x^2 - 8x + 7 = 0\).

Step4: Solve the quadratic equation

Factor the quadratic: \((x - 1)(x - 7)=0\). So the solutions are \(x = 1\) or \(x = 7\).

Step5: Check the solutions

• For \(x = 1\): Substitute into the original equation, left side \(2+\sqrt{4(1)-3}=2 + 1=3\), right side \(1\). Since \(3

eq1\), \(x = 1\) is extraneous.

• For \(x = 7\): Substitute into the original equation, left side \(2+\sqrt{4(7)-3}=2+\sqrt{25}=2 + 5 = 7\), right side \(7\). So \(x = 7\) is a valid solution.

Answer:

\(x = 7\)