Solve for v, where v is a real number. √-v+6=v

To solve the equation \(\sqrt{-v + 6} = v\) where \(v\) is a real number, follow these steps:

### Step 1: Isolate the square root term
Start with the equation:

\[
\sqrt{-v + 6} = v
\]

### Step 2: Square both sides to eliminate the square root
Squaring both sides gives:

\[
(\sqrt{-v + 6})^2 = v^2
\]

This simplifies to:

\[
-v + 6 = v^2
\]

### Step 3: Rearrange the equation to set it equal to zero
Rearrange the equation to form a standard quadratic equation:

\[
v^2 + v - 6 = 0
\]

### Step 4: Solve the quadratic equation
This is a quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = 1\), and \(c = -6\). Use the quadratic formula:

\[
v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substituting the values:

\[
v = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)}
\]

\[
v = \frac{-1 \pm \sqrt{1 + 24}}{2}
\]

\[
v = \frac{-1 \pm \sqrt{25}}{2}
\]

\[
v = \frac{-1 \pm 5}{2}
\]

### Step 5: Calculate the solutions
This gives two possible solutions:

\[
v = \frac{-1 + 5}{2} = \frac{4}{2} = 2
\]

\[
v = \frac{-1 - 5}{2} = \frac{-6}{2} = -3
\]

### Step 6: Check the solutions in the original equation
Let's check both solutions in the original equation \(\sqrt{-v + 6} = v\).

- For \(v = 2\):

\[
\sqrt{-2 + 6} = \sqrt{4} = 2 \quad \text{(True)}
\]

- For \(v = -3\):

\[
\sqrt{-(-3) + 6} = \sqrt{3 + 6} = \sqrt{9} = 3 \quad \text{(False, since \(-3 \neq 3\))}
\]

### Conclusion:
The only solution that satisfies the original equation is:

\[
v = 2
\]

Would you like further details or have any other questions?

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Here are 5 related questions you might find useful:
1. How do you solve quadratic equations using the quadratic formula?
2. What are the methods for solving square root equations?
3. What is the significance of checking solutions in the original equation?
4. How do you determine the domain of a square root function?
5. Can a quadratic equation have no real solutions?

**Tip:** Always check each solution in the original equation, especially when dealing with square roots or quadratic equations, as extraneous solutions can arise.

Learn how to solve the equation √(-v + 6) = v with a detailed step-by-step solution. This problem covers key algebraic concepts, including quadratic equations and square roots. Suitable for students in Grades 9-12.

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