Question

matk311_mathematics iii semester a_2026-080220207-integ math 3a
polynomials, rational, and radical relationships: part three
which statement about \\(\sqrt{x} - \sqrt{x - 5} = 1\\) is true?
\\(x = 4\\) is an extraneous solution.
\\(x = 9\\) is an extraneous solution.
\\(x = 9\\) is a true solution.
\\(x = 4\\) is a true solution.

Explanation:

Step1: Isolate one radical

Start with the equation \(\sqrt{x} - \sqrt{x - 5} = 1\). Isolate \(\sqrt{x}\) by adding \(\sqrt{x - 5}\) to both sides: \(\sqrt{x} = 1 + \sqrt{x - 5}\).

Step2: Square both sides

Square both sides to eliminate the square roots: \((\sqrt{x})^2 = (1 + \sqrt{x - 5})^2\). This simplifies to \(x = 1 + 2\sqrt{x - 5} + (x - 5)\).

Step3: Simplify the equation

Simplify the right - hand side: \(x = 1 + 2\sqrt{x - 5}+x - 5\). Subtract \(x\) from both sides: \(0=1 - 5+2\sqrt{x - 5}\), which further simplifies to \(0=-4 + 2\sqrt{x - 5}\).

Step4: Solve for the radical

Add 4 to both sides: \(4 = 2\sqrt{x - 5}\). Divide both sides by 2: \(2=\sqrt{x - 5}\).

Step5: Square both sides again

Square both sides: \(2^2=(\sqrt{x - 5})^2\), so \(4=x - 5\). Add 5 to both sides: \(x = 9\).

Step6: Check for extraneous solutions

Now we need to check if \(x = 9\) is a valid solution by plugging it back into the original equation.
Left - hand side: \(\sqrt{9}-\sqrt{9 - 5}=3-\sqrt{4}=3 - 2 = 1\).
Right - hand side: \(1\).
Since the left - hand side equals the right - hand side when \(x = 9\), \(x = 9\) is a true solution.
Now check \(x = 4\):
Left - hand side: \(\sqrt{4}-\sqrt{4 - 5}=2-\sqrt{-1}\). But the square root of a negative number is not a real number in the set of real numbers, so \(x = 4\) does not satisfy the original equation (it is an extraneous solution).

Answer:

\(x = 9\) is a true solution.