Question

identify the solution(s) to \\(sqrt{25 - 4x} - x + 1 = 0\\)
hint: isolate the radical, square, solve, then check domain \\(x \geq 1\\)
skill sheet solving radical equations steps, p. 3

a. the solution is x=4

b. the solutions are x=4 and x=-6

c. the solution is x=-6

d. the solution set is {∅}

Explanation:

Step1: Isolate the radical

Start with the equation \(\sqrt{25 - 4x}-x + 1 = 0\). Isolate the square root term:
\(\sqrt{25 - 4x}=x - 1\)

Step2: Square both sides

Square both sides to eliminate the square root. Remember that \((\sqrt{a})^2=a\) for \(a\geq0\):
\((\sqrt{25 - 4x})^2=(x - 1)^2\)
\(25-4x=x^{2}-2x + 1\)

Step3: Rearrange into quadratic equation

Bring all terms to one side to form a quadratic equation:
\(x^{2}-2x + 1-25 + 4x = 0\)
\(x^{2}+2x-24 = 0\)

Step4: Solve the quadratic equation

Factor the quadratic equation \(x^{2}+2x - 24=0\). We need two numbers that multiply to \(- 24\) and add to \(2\). The numbers are \(6\) and \(-4\):
\((x + 6)(x - 4)=0\)
Set each factor equal to zero:
\(x+6 = 0\) or \(x - 4=0\)
So \(x=-6\) or \(x = 4\)

Step5: Check the domain and solutions

First, the domain of the square root \(\sqrt{25-4x}\) requires \(25-4x\geq0\), so \(x\leq\frac{25}{4}=6.25\). Also, from the isolated radical \(\sqrt{25 - 4x}=x - 1\), the right - hand side \(x - 1\) must be non - negative (since the square root is non - negative), so \(x-1\geq0\) or \(x\geq1\).

• For \(x = 4\):

Check the original equation: \(\sqrt{25-4\times4}-4 + 1=\sqrt{25 - 16}-3=\sqrt{9}-3=3 - 3=0\). So \(x = 4\) is a valid solution.

• For \(x=-6\):

Check the original equation: \(\sqrt{25-4\times(-6)}-(-6)+1=\sqrt{25 + 24}+7=\sqrt{49}+7=7 + 7=14
eq0\). Also, \(x=-6\) does not satisfy \(x\geq1\), so \(x=-6\) is an extraneous solution.

Answer:

A. The solution is \(x = 4\)