Question

6.) \sqrt{3n + 10} = n + 4

Explanation:

Step1: Square both sides to eliminate the square root

To get rid of the square root, we square both sides of the equation \(\sqrt{3n + 10}=n + 4\). Squaring the left side gives \(3n + 10\), and squaring the right side \((n + 4)^2\) expands to \(n^2+8n + 16\) using the formula \((a + b)^2=a^2+2ab + b^2\) where \(a = n\) and \(b = 4\). So we have the equation:
\(3n+10=n^{2}+8n + 16\)

Step2: Rearrange into standard quadratic form

Subtract \(3n\) and \(10\) from both sides to set the equation to zero:
\(n^{2}+8n + 16-3n - 10=0\)
Simplify the like terms:
\(n^{2}+5n + 6=0\)

Step3: Factor the quadratic equation

We need two numbers that multiply to \(6\) and add up to \(5\). The numbers \(2\) and \(3\) work. So we can factor the quadratic as:
\((n + 2)(n+3)=0\)

Step4: Solve for n using the zero - product property

If \((n + 2)(n + 3)=0\), then either \(n+2 = 0\) or \(n + 3=0\).

• If \(n+2=0\), then \(n=-2\).
• If \(n + 3=0\), then \(n=-3\).

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides of an equation can introduce extraneous solutions.

• For \(n=-2\):

Left side: \(\sqrt{3(-2)+10}=\sqrt{-6 + 10}=\sqrt{4}=2\)
Right side: \(-2 + 4=2\)
Since \(2 = 2\), \(n=-2\) is a valid solution.

• For \(n=-3\):

Left side: \(\sqrt{3(-3)+10}=\sqrt{-9 + 10}=\sqrt{1}=1\)
Right side: \(-3 + 4=1\)
Since \(1=1\), \(n=-3\) is also a valid solution.

Answer:

The solutions are \(n=-2\) and \(n=-3\)