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6.2 hw - solving radicals
score: 14/20 answered: 7/10
question 8
solve the following radical equation. enter integers or reduced fractions as your answer. do not use decimals. if there is more than one solution, separate the solutions with a comma. if there is no solutio use dne.
solve: $3x = \sqrt{36x + 108}$
solutions: $x = \square$
extraneous solutions: $x = \square$

Explanation:

Step1: Square both sides to eliminate the square root.

To solve the radical equation \(3x = \sqrt{36x + 108}\), we first square both sides. Squaring the left side gives \((3x)^2 = 9x^2\), and squaring the right side gives \((\sqrt{36x + 108})^2 = 36x + 108\). So we have the equation:
\[9x^2 = 36x + 108\]

Step2: Rearrange the equation to standard quadratic form.

Subtract \(36x\) and \(108\) from both sides to get:
\[9x^2 - 36x - 108 = 0\]
We can simplify this equation by dividing all terms by \(9\):
\[x^2 - 4x - 12 = 0\]

Step3: Factor the quadratic equation.

We need to find two numbers that multiply to \(-12\) and add up to \(-4\). The numbers are \(-6\) and \(2\). So we can factor the quadratic as:
\[(x - 6)(x + 2) = 0\]

Step4: Solve for \(x\) using the zero - product property.

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:

• If \(x - 6 = 0\), then \(x = 6\).
• If \(x + 2 = 0\), then \(x=-2\).

Step5: Check for extraneous solutions.

We need to check these solutions in the original equation \(3x=\sqrt{36x + 108}\) because squaring both sides can introduce extraneous solutions.

• For \(x = 6\):

Left side: \(3x=3\times6 = 18\)
Right side: \(\sqrt{36\times6+108}=\sqrt{216 + 108}=\sqrt{324}=18\)
Since \(18 = 18\), \(x = 6\) is a valid solution.

• For \(x=-2\):

Left side: \(3x=3\times(-2)=-6\)
Right side: \(\sqrt{36\times(-2)+108}=\sqrt{-72 + 108}=\sqrt{36}=6\)
Since \(-6
eq6\), \(x=-2\) is an extraneous solution.

Answer:

Solutions: \(x = 6\)
Extraneous Solutions: \(x=-2\)