Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
2x + root ( x + 6)= 9
x =

To solve the equation \( 2x + \sqrt{x + 6} = 9 \), let's follow these steps:

### Step 1: Isolate the square root term
First, isolate the square root term by subtracting \(2x\) from both sides of the equation:
\[
\sqrt{x + 6} = 9 - 2x
\]

### Step 2: Square both sides
Square both sides of the equation to eliminate the square root:
\[
(\sqrt{x + 6})^2 = (9 - 2x)^2
\]
\[
x + 6 = (9 - 2x)^2
\]

### Step 3: Expand the square on the right-hand side
Now, expand \( (9 - 2x)^2 \):
\[
x + 6 = (9 - 2x)(9 - 2x)
\]
\[
x + 6 = 81 - 36x + 4x^2
\]

### Step 4: Rearrange the equation
Rearrange the equation into a standard quadratic form:
\[
4x^2 - 36x - x + 81 - 6 = 0
\]
\[
4x^2 - 37x + 75 = 0
\]

### Step 5: Solve the quadratic equation
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -37\), and \(c = 75\):

\[
x = \frac{-(-37) \pm \sqrt{(-37)^2 - 4(4)(75)}}{2(4)}
\]
\[
x = \frac{37 \pm \sqrt{1369 - 1200}}{8}
\]
\[
x = \frac{37 \pm \sqrt{169}}{8}
\]
\[
x = \frac{37 \pm 13}{8}
\]

### Step 6: Find the two possible values of \(x\)
\[
x_1 = \frac{37 + 13}{8} = \frac{50}{8} = 6.25
\]
\[
x_2 = \frac{37 - 13}{8} = \frac{24}{8} = 3
\]

### Step 7: Check the solutions in the original equation
Now, substitute these values back into the original equation to check if they are valid:

1. For \(x = 6.25\):
   \[
   2(6.25) + \sqrt{6.25 + 6} = 12.5 + \sqrt{12.25} \neq 9
   \]
   This does not satisfy the original equation, so \(x = 6.25\) is not a solution.

2. For \(x = 3\):
   \[
   2(3) + \sqrt{3 + 6} = 6 + \sqrt{9} = 6 + 3 = 9
   \]
   This satisfies the original equation, so \(x = 3\) is a valid solution.

### Final Answer:
The only real solution is:
\[
x = 3
\]

Would you like a further breakdown or have any questions?

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1. What is the general method for solving radical equations?
2. How can we determine whether a quadratic equation has real solutions?
3. What are the potential pitfalls when squaring both sides of an equation?
4. How do you simplify expressions with square roots?
5. What are common methods for checking solutions to algebraic equations?
6. How do you factor quadratic equations?
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8. When is it necessary to check for extraneous solutions?

**Tip:** Always check your solutions in the original equation, especially when dealing with square roots or other powers.

Explore how to find real solutions for the equation 2x + sqrt(x + 6) = 9 with a step-by-step solution. This problem involves quadratic equations and square roots. Suitable for high school students studying algebra.

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