Question

solve for x. check your solutions.\\(\sqrt{2x + 1} + 29 = 4\\)\\(x = 2\\) \\(x = 4\\) \\(x = 2, 4\\) no solution

Explanation:

Step1: Isolate the square root term

Subtract 29 from both sides of the equation $\sqrt{2x + 1}+29 = 4$.
$\sqrt{2x + 1}=4 - 29$
$\sqrt{2x + 1}=- 25$

Step2: Analyze the square root's range

The square root function $y = \sqrt{a}$ (where $a = 2x + 1$ here) has a range of $y\geq0$, meaning $\sqrt{2x + 1}\geq0$ for all real - valued $x$ that make the expression under the square root defined. But we have $\sqrt{2x + 1}=-25$, and $-25<0$, which is a contradiction. So, there is no real - valued solution for $x$.

Answer:

No Solution