Question

which statement about $sqrt{2x + 3} = sqrt{2x} + 3$ is true?$x = 2$ is a true solution$x = 2$ is an extraneous solution$x = \frac{1}{2}$ is an extraneous solution$x = \frac{1}{2}$ is a true solution

Explanation:

Step1: Square both sides to eliminate roots

$$(\sqrt{2x + 3})^2 = (\sqrt{2x} + 3)^2$$
$$2x + 3 = 2x + 6\sqrt{2x} + 9$$

Step2: Simplify the equation

$$2x + 3 - 2x - 9 = 6\sqrt{2x}$$
$$-6 = 6\sqrt{2x}$$
$$-1 = \sqrt{2x}$$

Step3: Square both sides again

$$(-1)^2 = (\sqrt{2x})^2$$
$$1 = 2x$$
$$x = \frac{1}{2}$$

Step4: Verify the solution

Substitute $x=\frac{1}{2}$ into original equation:
Left side: $\sqrt{2\times\frac{1}{2}+3}=\sqrt{1+3}=\sqrt{4}=2$
Right side: $\sqrt{2\times\frac{1}{2}}+3=\sqrt{1}+3=1+3=4$
*Note: Correction: Re-check Step 2 algebra*

Step1 (Revised): Square both sides correctly

$$2x + 3 = 2x + 6\sqrt{2x} + 9$$

Step2 (Revised): Isolate the radical term

$$3 - 9 = 6\sqrt{2x}$$
$$-6 = 6\sqrt{2x}$$
$$\sqrt{2x} = -1$$

Step3 (Revised): Analyze radical validity

A square root cannot equal a negative number, so there is no real solution from this. However, re-test $x=\frac{1}{2}$:
Left: $\sqrt{2(\frac{1}{2})+3}=\sqrt{4}=2$
Right: $\sqrt{2(\frac{1}{2})}+3=1+3=4$
*Note: Correct approach: Let $y=\sqrt{2x}$, $y\geq0$
Equation: $\sqrt{y^2 + 3} = y + 3$
Square: $y^2+3=y^2+6y+9$
$0=6y+6$
$y=-1$, which is invalid. But re-express original equation correctly:
Wait, original equation: $\sqrt{2x+3}=\sqrt{2x}+3$
Let $t=\sqrt{2x}, t\geq0$, so $2x=t^2$, $2x+3=t^2+3$
Equation: $\sqrt{t^2+3}=t+3$
Square both sides: $t^2+3=t^2+6t+9$
$3=6t+9$
$6t=-6$
$t=-1$
Since $t=\sqrt{2x}\geq0$, $t=-1$ is invalid, so no solution? But test $x=2$:
Left: $\sqrt{4+3}=\sqrt{7}\approx2.645$
Right: $\sqrt{4}+3=2+3=5$, not equal.
Test $x=\frac{1}{2}$:
Left: $\sqrt{1+3}=2$
Right: $\sqrt{1}+3=4$, not equal.
Wait, error in initial test: Correct solving:
$\sqrt{2x+3}-\sqrt{2x}=3$
Rationalize: Multiply numerator and denominator by $\sqrt{2x+3}+\sqrt{2x}$:
$\frac{(2x+3)-2x}{\sqrt{2x+3}+\sqrt{2x}}=3$
$\frac{3}{\sqrt{2x+3}+\sqrt{2x}}=3$
$\sqrt{2x+3}+\sqrt{2x}=1$
Now we have system:
1. $\sqrt{2x+3}-\sqrt{2x}=3$
2. $\sqrt{2x+3}+\sqrt{2x}=1$
Add equations: $2\sqrt{2x+3}=4$ → $\sqrt{2x+3}=2$ → $2x+3=4$ → $2x=1$ → $x=\frac{1}{2}$
Subtract equations: $-2\sqrt{2x}=2$ → $\sqrt{2x}=-1$, which is impossible. But when we substitute $x=\frac{1}{2}$ into original equation, left=2, right=4, which is not equal. Wait, no—wait, the rationalization step created a contradiction, meaning no solution? But the options say $x=\frac{1}{2}$ is true? No, wait, no:
Wait, original equation: $\sqrt{2x+3}=\sqrt{2x}+3$
Square both sides: $2x+3=2x+6\sqrt{2x}+9$
Simplify: $0=6\sqrt{2x}+6$ → $\sqrt{2x}=-1$
This has no real solution, but if we ignore the domain and solve for x, we get $x=\frac{1}{2}$, which is extraneous? No, wait, no:
Wait, no—when we square, we can introduce extraneous solutions. So $x=\frac{1}{2}$ is extraneous? But that contradicts. Wait, no:
Wait, let's compute left and right for $x=\frac{1}{2}$:
Left: $\sqrt{2*(1/2)+3}=\sqrt{1+3}=2$
Right: $\sqrt{2*(1/2)}+3=\sqrt{1}+3=4$
2≠4, so $x=\frac{1}{2}$ is extraneous.
Wait, what about x=2:
Left: $\sqrt{4+3}=\sqrt{7}≈2.645$
Right: $\sqrt{4}+3=2+3=5$
2.645≠5, so x=2 is extraneous.
Wait, but that can't be. Wait, no—wait, is there any solution?
Let $f(x)=\sqrt{2x+3}-\sqrt{2x}-3$
For $x\geq0$, $\sqrt{2x+3}\leq\sqrt{2x}+3$?
$\sqrt{2x+3}\leq\sqrt{2x}+3$
Square: $2x+3\leq2x+6\sqrt{2x}+9$
$3\leq6\sqrt{2x}+9$
$-6\leq6\sqrt{2x}$
$-1\leq\sqrt{2x}$, which is always true for x≥0. But equality holds only when $\sqrt{2x}=-1$, which is impossible. So there is no real solution. But the options are:
1. x=2 is true
2. x=2 is extraneous
3. x=1/2 is extraneous
4. x=1/2 is true
Wait, when we solve, we get x=1/2 as a candidate, but it doesn't satisfy the original equation, so it's extraneous. x=2 is not a candidate from solving. So the correct answer is $x=\frac{1}{2}$ is an extraneous solution? Wait no, wait:
Wait, let's re-solve correctly:
$\sqrt{2x+3}=\sqrt{2x}+3$
Let $u=\sqrt{2x}$, u≥0
$\sqrt{u^2+3}=u+3$
Square: $u^2+3=u^2+6u+9$
$3=6u+9$
$6u=-6$
$u=-1$
Since u≥0, no solution. But if we solve for x from u=-1, $u=\sqrt{2x}=-1$ → $2x=1$ → x=1/2, which is a candidate, but u=-1 is invalid, so x=1/2 is extraneous.
Yes, that's correct. So x=1/2 is extraneous. Wait but earlier when I thought x=1/2 is true, that was wrong, because left≠right.
Wait, no: left=2, right=4, 2≠4, so x=1/2 is extraneous.
So the correct answer is $x=\frac{1}{2}$ is an extraneous solution.

Final Correct Answer:

$x = \frac{1}{2}$ is an extraneous solution.

Explanation:

Step1: Substitute variable for radical

Let $u=\sqrt{2x}$ where $u\geq0$. Equation becomes:
$$\sqrt{u^2 + 3} = u + 3$$

Step2: Square both sides

$$u^2 + 3 = (u + 3)^2$$
$$u^2 + 3 = u^2 + 6u + 9$$

Step3: Simplify to solve for u

$$3 = 6u + 9$$
$$6u = -6$$
$$u = -1$$

Step4: Convert back to x

$$\sqrt{2x} = -1$$
Square both sides: $2x = 1$ → $x=\frac{1}{2}$

Step5: Verify solution

Substitute $x=\frac{1}{2}$ into original equation:
Left: $\sqrt{2(\frac{1}{2})+3}=\sqrt{4}=2$
Right: $\sqrt{2(\frac{1}{2})}+3=1+3=4$
Since $2
eq4$, $x=\frac{1}{2}$ is extraneous.

Step1: Substitute variable for radical

Let $u=\sqrt{2x}$ where $u\geq0$. Equation becomes:
$$\sqrt{u^2 + 3} = u + 3$$

Step2: Square both sides

$$u^2 + 3 = (u + 3)^2$$
$$u^2 + 3 = u^2 + 6u + 9$$

Step3: Simplify to solve for u

$$3 = 6u + 9$$
$$6u = -6$$
$$u = -1$$

Step4: Convert back to x

$$\sqrt{2x} = -1$$
Square both sides: $2x = 1$ → $x=\frac{1}{2}$

Step5: Verify solution

Substitute $x=\frac{1}{2}$ into original equation:
Left: $\sqrt{2(\frac{1}{2})+3}=\sqrt{4}=2$
Right: $\sqrt{2(\frac{1}{2})}+3=1+3=4$
Since $2
eq4$, $x=\frac{1}{2}$ is extraneous.

Answer:

$x = \frac{1}{2}$ is a true solution.