Question

solve for v, where v is a real number.
v - 3 = \sqrt{-5v + 15}
if there is more than one solution, separate them with commas.
if there is no solution, click on
o solution\.
v =

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \( v - 3=\sqrt{-5v + 15} \), we first square both sides. Squaring the left side \((v - 3)^2\) gives \(v^{2}-6v + 9\), and squaring the right side \((\sqrt{-5v + 15})^2\) gives \(-5v + 15\). So we get the quadratic equation:

$$v^{2}-6v + 9=-5v + 15$$

Step2: Rearrange the quadratic equation

Move all terms to the left side of the equation to set it to zero. Add \(5v\) to both sides and subtract \(15\) from both sides:

$$v^{2}-6v+5v + 9 - 15 = 0$$

Simplify the like terms:

$$v^{2}-v - 6=0$$

Step3: Factor the quadratic equation

We factor the quadratic \(v^{2}-v - 6\). We need two numbers that multiply to \(-6\) and add up to \(-1\). The numbers are \(-3\) and \(2\). So we can factor the quadratic as:

$$(v - 3)(v + 2)=0$$

Step4: Solve for \(v\) 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 \(v - 3=0\), then \(v=3\).
• If \(v + 2=0\), then \(v=-2\).

Step5: Check the solutions in the original equation

• For \(v = 3\):

Substitute \(v = 3\) into the original equation. The left side is \(3-3 = 0\), and the right side is \(\sqrt{-5\times3 + 15}=\sqrt{-15 + 15}=\sqrt{0}=0\). So \(v = 3\) is a valid solution.

• For \(v=-2\):

Substitute \(v=-2\) into the original equation. The left side is \(-2-3=-5\), and the right side is \(\sqrt{-5\times(-2)+15}=\sqrt{10 + 15}=\sqrt{25}=5\). Since \(-5
eq5\), \(v=-2\) is not a valid solution.

Answer:

\(3\)