Question

matk31l_mathematics iii semester a_2026-080220207-integ math 3a
polynomials, rational, and radical relationships: part three
what is the solution of \\(\sqrt{x - 4} + 5 = 2\\)?
\\(x = -17\\)
\\(x = 13\\)
\\(x = 53\\)
no solution

Explanation:

Step1: Isolate the radical term

Subtract 5 from both sides of the equation $\sqrt{x - 4} + 5 = 2$.
We get $\sqrt{x - 4} = 2 - 5 = -3$.

Step2: Analyze the radical's nature

The square root function $\sqrt{a}$ (where $a=x - 4$ here) is defined such that $\sqrt{a} \geq 0$ for all real numbers $a\geq0$. But we have $\sqrt{x - 4} = -3$, and the square root of a number cannot be negative in the set of real numbers. So, there is no real solution.

Answer:

no solution