If f(x)=`sqrt(2x+3)` and g(x)=`x^2` , for what value (s) of x does f(g(x))=g(f(x))?

Options

A. -0.55
B. 0.46
C. 5.45
D. -0.55 and 5.45

Explanation

To solve the equation \( f(g(x)) = g(f(x)) \) where \( f(x) = \sqrt{2x + 3} \) and \( g(x) = x^2 \), we will follow these steps:

### Step 1: Find \( f(g(x)) \)
Substituting \( g(x) = x^2 \) into \( f(x) \):
\[ f(g(x)) = f(x^2) = \sqrt{2(x^2) + 3} = \sqrt{2x^2 + 3} \]

### Step 2: Find \( g(f(x)) \)
Substituting \( f(x) = \sqrt{2x + 3} \) into \( g(x) \):
\[ g(f(x)) = g(\sqrt{2x + 3}) = (\sqrt{2x + 3})^2 = 2x + 3 \]

### Step 3: Set the two expressions equal to each other
Now we have:
\[ \sqrt{2x^2 + 3} = 2x + 3 \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 2x^2 + 3 = (2x + 3)^2 \]
Expanding the right side:
\[ (2x + 3)^2 = 4x^2 + 12x + 9 \]
So we have:
\[ 2x^2 + 3 = 4x^2 + 12x + 9 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 0 = 4x^2 + 12x + 9 - 2x^2 - 3 \]
This simplifies to:
\[ 0 = 2x^2 + 12x + 6 \]

### Step 6: Factor out the common term
Factoring out 2:
\[ 0 = 2(x^2 + 6x + 3) \]
Thus, we can ignore the factor of 2:
\[ 0 = x^2 + 6x + 3 \]

### Step 7: Use the quadratic formula to find the roots
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 6, c = 3 \):
\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{-6 \pm \sqrt{36 - 12}}{2} = \frac{-6 \pm \sqrt{24}}{2} \]
Simplifying further:
\[ \sqrt{24} = 2\sqrt{6} \]
Thus:
\[ x = \frac{-6 \pm 2\sqrt{6}}{2} = -3 \pm \sqrt{6} \]

### Step 8: Determine valid solutions based on the domain of \( f(x) \)
Since \( f(x) = \sqrt{2x + 3} \), we need \( 2x + 3 \geq 0 \):
\[ 2x \geq -3 \implies x \geq -\frac{3}{2} \]

### Step 9: Evaluate the solutions
The two potential solutions are:
\[ x_1 = -3 + \sqrt{6} \quad \text{and} \quad x_2 = -3 - \sqrt{6} \]
Calculating \( x_1 \):
\[ -3 + \sqrt{6} \approx -3 + 2.45 \approx -0.55 \]
Calculating \( x_2 \):
\[ -3 - \sqrt{6} \approx -3 - 2.45 \approx -5.45 \]
Since \( x_2 < -\frac{3}{2} \), it is not valid. Thus, the only valid solution is:
\[ x = -3 + \sqrt{6} \]

### Final Answer
The value of \( x \) for which \( f(g(x)) = g(f(x)) \) is:
\[ x = -3 + \sqrt{6} \]