Solve the radical equation.

StartRoot 2 x plus 3 EndRoot2x+3plus+StartRoot x minus 2 EndRootx−2equals=2

We are given the radical equation:

\[
\sqrt{2x + 3} + \sqrt{x - 2} = 2
\]

### Step 1: Isolate one of the radicals
Let's isolate one of the square roots. We can start by isolating \(\sqrt{2x + 3}\) on one side of the equation:

\[
\sqrt{2x + 3} = 2 - \sqrt{x - 2}
\]

### Step 2: Square both sides
To eliminate the square root on the left side, square both sides of the equation:

\[
(\sqrt{2x + 3})^2 = (2 - \sqrt{x - 2})^2
\]

This simplifies to:

\[
2x + 3 = (2 - \sqrt{x - 2})^2
\]

Now, expand the right-hand side:

\[
2x + 3 = (2)^2 - 2(2)(\sqrt{x - 2}) + (\sqrt{x - 2})^2
\]

\[
2x + 3 = 4 - 4\sqrt{x - 2} + (x - 2)
\]

Simplifying further:

\[
2x + 3 = 4 + x - 2 - 4\sqrt{x - 2}
\]

\[
2x + 3 = x + 2 - 4\sqrt{x - 2}
\]

### Step 3: Isolate the remaining radical
Now, isolate the radical term:

\[
2x + 3 - x - 2 = -4\sqrt{x - 2}
\]

\[
x + 1 = -4\sqrt{x - 2}
\]

### Step 4: Square both sides again
To eliminate the remaining square root, square both sides again:

\[
(x + 1)^2 = (-4\sqrt{x - 2})^2
\]

\[
(x + 1)^2 = 16(x - 2)
\]

Expand both sides:

\[
x^2 + 2x + 1 = 16x - 32
\]

### Step 5: Bring everything to one side and simplify
Now, bring all terms to one side of the equation:

\[
x^2 + 2x + 1 - 16x + 32 = 0
\]

\[
x^2 - 14x + 33 = 0
\]

### Step 6: Solve the quadratic equation
Now, solve the quadratic equation \(x^2 - 14x + 33 = 0\) using the quadratic formula:

\[
x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(33)}}{2(1)}
\]

\[
x = \frac{14 \pm \sqrt{196 - 132}}{2}
\]

\[
x = \frac{14 \pm \sqrt{64}}{2}
\]

\[
x = \frac{14 \pm 8}{2}
\]

So, we have two possible solutions:

\[
x = \frac{14 + 8}{2} = \frac{22}{2} = 11
\]
\[
x = \frac{14 - 8}{2} = \frac{6}{2} = 3
\]

### Step 7: Check for extraneous solutions
We must check both values in the original equation to see if they work.

1. For \(x = 11\):

\[
\sqrt{2(11) + 3} + \sqrt{11 - 2} = \sqrt{22 + 3} + \sqrt{9} = \sqrt{25} + 3 = 5 + 3 = 8 \neq 2
\]

So, \(x = 11\) is **not a solution**.

2. For \(x = 3\):

\[
\sqrt{2(3) + 3} + \sqrt{3 - 2} = \sqrt{6 + 3} + \sqrt{1} = \sqrt{9} + 1 = 3 + 1 = 4 \neq 2
\]

So, \(x = 3\) is also **not a solution**.

### Conclusion:
This radical equation has **no real solutions**.

Would you like further details or have any questions?

### Related Questions:
1. How do you solve equations involving multiple square roots?
2. What are the common types of extraneous solutions in radical equations?
3. How does squaring both sides of an equation affect possible solutions?
4. What is the quadratic formula, and how is it derived?
5. How do you check for extraneous solutions in other types of equations?

### Tip:
Always check for extraneous solutions when solving radical equations by substituting potential solutions back into the original equation.

Solve the radical equation: sqrt(2x + 3) + sqrt(x - 2) = 2

Explore how to solve the radical equation sqrt(2x + 3) + sqrt(x - 2) = 2 with a step-by-step solution. This algebra problem involves isolating square roots, squaring both sides, and solving a quadratic equation. Find out why the equation has no real solutions. Suitable for high school students in Grades 10-12.

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