Question

solve the radical equation. check all proposed solutions.
\\(\sqrt{15 - 2x} = x\\)

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

\\(\bigcirc\\) a. the solution set is \\(\\{\\}\\).
(simplify your answer. use a comma to separate answers as needed.)
\\(\bigcirc\\) b. the solution set is the empty set.

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{15 - 2x}=x\), we square both sides. Squaring the left side \((\sqrt{15 - 2x})^2\) gives \(15 - 2x\), and squaring the right side \(x^2\) gives \(x^2\). So we get the quadratic equation:
\(15 - 2x=x^2\)

Step2: Rearrange the quadratic equation

Rearrange the equation \(15 - 2x=x^2\) into standard quadratic form \(ax^2+bx + c = 0\). We move all terms to one side:
\(x^2+2x - 15=0\)

Step3: Factor the quadratic equation

Factor the quadratic \(x^2 + 2x-15\). We need two numbers that multiply to \(- 15\) and add to \(2\). The numbers are \(5\) and \(-3\). So the factored form is:
\((x + 5)(x - 3)=0\)

Step4: Solve for x using the zero - product property

Using the zero - product property, if \((x + 5)(x - 3)=0\), then either \(x+5 = 0\) or \(x - 3=0\).

• If \(x+5=0\), then \(x=-5\).
• If \(x - 3=0\), then \(x = 3\).

Step5: Check the proposed solutions

Check \(x=-5\):

Substitute \(x = - 5\) into the original equation \(\sqrt{15-2x}=x\).
Left side: \(\sqrt{15-2\times(-5)}=\sqrt{15 + 10}=\sqrt{25}=5\)
Right side: \(x=-5\)
Since \(5
eq - 5\), \(x=-5\) is an extraneous solution.

Check \(x = 3\):

Substitute \(x = 3\) into the original equation \(\sqrt{15-2x}=x\).
Left side: \(\sqrt{15-2\times3}=\sqrt{15 - 6}=\sqrt{9}=3\)
Right side: \(x = 3\)
Since the left side equals the right side when \(x = 3\), \(x = 3\) is a valid solution.

Answer:

A. The solution set is \(\{3\}\)