Question

\\\sqrt{x} + 7 = 2\\

Explanation:

🆕 New Concept Discovered: Solving Radical Equations
Isolate the root, square both sides, and check for extraneous solutions.

Step 1: Isolate the radical term

To solve for \(x\), we first want to get the square root term \(\sqrt{x}\) by itself on one side of the equation. We do this by subtracting \(7\) from both sides:

$$ \sqrt{x} + 7 - 7 = 2 - 7 $$

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

Step 2: Analyze the equation

A principal square root, \(\sqrt{x}\), represents the non-negative square root of a number. By definition, it can never produce a negative value:

$$ \sqrt{x} \ge 0 \quad \text{for all real } x \ge 0 $$

Since our equation states that \(\sqrt{x} = -5\), and a square root cannot equal a negative number, there is no real number \(x\) that satisfies this equation.

Step 3: Verify by squaring (optional check)

If we proceed by squaring both sides to eliminate the radical:

$$ (\sqrt{x})^2 = (-5)^2 $$

$$ x = 25 $$

Now, we must substitute \(x = 25\) back into the original equation to check for extraneous solutions:

$$ \sqrt{25} + 7 \stackrel{?}{=} 2 $$

$$ 5 + 7 \stackrel{?}{=} 2 $$

$$ 12 eq 2 $$

Since \(x = 25\) does not satisfy the original equation, it is an extraneous solution. Therefore, the equation has no solution.

Answer:

No solution (or no real solution)