Question

solve \\(\sqrt{2x - 1} + 5 = 2\\)

Explanation:

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

Step 1: Isolate the radical term

To solve the equation, we first want to get the square root term by itself on one side of the equals sign. We do this by subtracting \(5\) from both sides of the equation:

$$ \sqrt{2x - 1} + 5 = 2 $$

$$ \sqrt{2x - 1} = 2 - 5 $$

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

Step 2: Analyze the isolated equation

Now we look at the equation:

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

In real numbers, the principal square root of any expression, \(\sqrt{\text{expression}}\), is always non-negative (greater than or equal to \(0\)).

Since a square root cannot produce a negative value, it is impossible for \(\sqrt{2x - 1}\) to equal \(-3\).

Step 3: Verify algebraically (optional check)

If we attempted to solve by squaring both sides:

$$ (\sqrt{2x - 1})^2 = (-3)^2 $$

$$ 2x - 1 = 9 $$

$$ 2x = 10 $$

$$ x = 5 $$

If we plug \(x = 5\) back into our original equation to check our work:

$$ \sqrt{2(5) - 1} + 5 = \sqrt{9} + 5 = 3 + 5 = 8 eq 2 $$

Because \(8
eq 2\), the value \(x = 5\) is an extraneous solution (a false solution introduced by squaring). Therefore, there are no real values of \(x\) that satisfy the equation.

Answer:

No solution (or no real solution)