Question

solve for v, where v is a real number.

v + 2 = \sqrt{-v + 28}

if there is more than one solution, separate them with commas.
if there is no solution, click on
o solution\.

v = \square

Explanation:

Step1: Square both sides to eliminate square root

To solve the equation \( v + 2=\sqrt{-v + 28} \), we first square both sides. Squaring the left side \((v + 2)\) gives \((v + 2)^2=v^{2}+4v + 4\), and squaring the right side \(\sqrt{-v + 28}\) gives \(-v + 28\). So we get the equation:
\(v^{2}+4v + 4=-v + 28\)

Step2: Rearrange into standard quadratic form

Move all terms to the left side of the equation to set it to zero. Add \(v\) to both sides and subtract 28 from both sides:
\(v^{2}+4v + v+4 - 28 = 0\)
Simplify the like terms:
\(v^{2}+5v - 24 = 0\)

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(-24\) and add up to \(5\). The numbers are \(8\) and \(-3\) because \(8\times(-3)=-24\) and \(8+(-3) = 5\). So we can factor the quadratic as:
\((v + 8)(v - 3)=0\)

Step4: Solve for v using zero - product property

Set each factor equal to zero:

• For \(v + 8=0\), we get \(v=-8\)
• For \(v - 3=0\), we get \(v = 3\)

Step5: Check for extraneous solutions

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

• Check \(v=-8\):

Left side: \(v + 2=-8 + 2=-6\)
Right side: \(\sqrt{-(-8)+28}=\sqrt{8 + 28}=\sqrt{36}=6\)
Since \(-6
eq6\), \(v = - 8\) is an extraneous solution.

• Check \(v = 3\):

Left side: \(v+2=3 + 2 = 5\)
Right side: \(\sqrt{-3 + 28}=\sqrt{25}=5\)
Since \(5 = 5\), \(v = 3\) is a valid solution.

Answer:

\(3\)