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Question
homework 10: section 2.6
score: 0/180 answered: 0/18
question 6
solve. list all answers separated by commas.
\\(\sqrt{6x + 16} = x\\)
Step1: Square both sides to eliminate the square root
To solve the equation \(\sqrt{6x + 16}=x\), we first square both sides of the equation. Squaring the left side \(\sqrt{6x + 16}\) gives \(6x + 16\), and squaring the right side \(x\) gives \(x^{2}\). So we get the equation:
\[6x + 16=x^{2}\]
Step2: Rearrange into standard quadratic form
We rearrange the equation \(6x + 16=x^{2}\) to the standard quadratic form \(ax^{2}+bx + c = 0\). Subtract \(6x\) and \(16\) from both sides:
\[x^{2}-6x - 16=0\]
Step3: Factor the quadratic equation
We factor the quadratic equation \(x^{2}-6x - 16 = 0\). We need two numbers that multiply to \(- 16\) and add up to \(-6\). The numbers are \(-8\) and \(2\) because \((-8)\times2=-16\) and \(-8 + 2=-6\). So the factored form is:
\[(x - 8)(x+2)=0\]
Step4: Solve for x using the zero - product property
Using the zero - product property, if \((x - 8)(x + 2)=0\), then either \(x-8 = 0\) or \(x + 2=0\).
- If \(x-8=0\), then \(x = 8\).
- If \(x + 2=0\), then \(x=-2\).
Step5: Check for extraneous solutions
We need to check these solutions in the original equation \(\sqrt{6x + 16}=x\) because squaring both sides can introduce extraneous solutions.
- For \(x = 8\):
Left side: \(\sqrt{6\times8+16}=\sqrt{48 + 16}=\sqrt{64}=8\)
Right side: \(8\)
Since the left side equals the right side, \(x = 8\) is a valid solution.
- For \(x=-2\):
Left side: \(\sqrt{6\times(-2)+16}=\sqrt{-12 + 16}=\sqrt{4}=2\)
Right side: \(-2\)
Since \(2
eq - 2\), \(x=-2\) is an extraneous solution and we discard it.
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