Solving Radical Equations: Finding the Value of x

The question asks us to find the value of x in the given radical equation: $\sqrt{2} + \sqrt{x} = \sqrt{3}$. To solve for x, we need to isolate the term with $\sqrt{x}$ and then square both sides of the equation to eliminate the radical.

Steps to Solve the Radical Equation

Here's a step-by-step process to solve for x:

1. Isolate the radical term: We need to get $\sqrt{x}$ by itself on one side of the equation. We can do this by subtracting $\sqrt{2}$ from both sides of the equation $\sqrt{2} + \sqrt{x} = \sqrt{3}$.
2. The equation becomes: $\sqrt{x} = \sqrt{3} - \sqrt{2}$.
3. Square both sides: To get rid of the square root on x, we square both sides of the equation. Squaring the left side gives us x. Squaring the right side involves squaring the expression $(\sqrt{3} - \sqrt{2})$.
4. So, $(\sqrt{x})^2 = (\sqrt{3} - \sqrt{2})^2$.
5. This simplifies to: $x = (\sqrt{3} - \sqrt{2})^2$.

Expanding the Squared Term

Now, we need to expand $(\sqrt{3} - \sqrt{2})^2$. We can use the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$. In this case, $a = \sqrt{3}$ and $b = \sqrt{2}$.

Applying the formula:

• $a^2 = (\sqrt{3})^2 = 3$
• $b^2 = (\sqrt{2})^2 = 2$
• $2ab = 2 \times \sqrt{3} \times \sqrt{2} = 2 \times \sqrt{3 \times 2} = 2\sqrt{6}$

So, $(\sqrt{3} - \sqrt{2})^2 = a^2 - 2ab + b^2 = 3 - 2\sqrt{6} + 2$.

Simplifying to Find x

Now we simplify the expression $3 - 2\sqrt{6} + 2$:

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

$x = 5 - 2\sqrt{6}$

Thus, the value of x is $5 - 2\sqrt{6}$.

Verifying the Solution with Options

We found that $x = 5 - 2\sqrt{6}$. Let's compare this with the given options:

Our calculated value $5 - 2\sqrt{6}$ matches Option 1.

Revision Table: Solving Radical Equations

Additional Information on Radical Equations

Radical equations are equations where the variable appears under a radical sign (like a square root). Solving them often involves isolating the radical and then raising both sides of the equation to the power equal to the index of the radical (e.g., squaring for a square root, cubing for a cube root).

• Isolating the Radical: It's usually best to get the radical term by itself on one side before squaring. If there are multiple radical terms, you might need to isolate one, square, and then repeat the process.
• Extraneous Solutions: Squaring both sides of an equation can sometimes introduce extraneous solutions. These are solutions that satisfy the squared equation but not the original equation. It is crucial to check your final answer(s) by substituting them back into the original radical equation to make sure they are valid. In this specific problem, since the original equation is $\sqrt{2} + \sqrt{x} = \sqrt{3}$, we must ensure $x \ge 0$. Our result $x = 5 - 2\sqrt{6}$. Since $5 = \sqrt{25}$ and $2\sqrt{6} = \sqrt{4 \times 6} = \sqrt{24}$, $5 - 2\sqrt{6} = \sqrt{25} - \sqrt{24}$ which is positive, so $x \ge 0$ is satisfied.
• Algebraic Identities: Remember common algebraic identities like $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$ as they are frequently used when squaring binomials involving radicals.