Question

solve the following radical equation. enter integers or reduced fractions as your answer. do not use decimals. if there is more than one solution, separate the solutions with a comma. if there is no solution, use dne.
solve: \\(\sqrt{6x + 13} = -7\\)
solutions: \\(x = \square\\)
extraneous solutions: \\(x = \square\\)

Explanation:

Step1: Analyze the square root property

The square root of a real number is always non - negative, i.e., for any real number \(y\), \(\sqrt{y}\geq0\) when \(y\) is in the domain of the square root function. In the equation \(\sqrt{6x + 13}=-7\), the left - hand side \(\sqrt{6x+13}\) represents a square root, so \(\sqrt{6x + 13}\geq0\) for all \(x\) in its domain. But the right - hand side is \(- 7\), which is negative.
However, let's try to solve the equation algebraically to check for extraneous solutions.
First, square both sides of the equation \(\sqrt{6x + 13}=-7\) to get rid of the square root.
\((\sqrt{6x + 13})^2=(-7)^2\)

Step2: Simplify both sides

Simplifying, we have \(6x+13 = 49\).
Then, subtract 13 from both sides: \(6x=49 - 13\)
\(6x=36\)
Divide both sides by 6: \(x = 6\)

Step3: Check for extraneous solutions

Now, substitute \(x = 6\) back into the original equation \(\sqrt{6x+13}\).
When \(x = 6\), \(6x+13=6\times6 + 13=36 + 13 = 49\), and \(\sqrt{49}=7
eq-7\). So \(x = 6\) is an extraneous solution, and the original equation has no solution.

Answer:

Solutions: DNE
Extraneous Solutions: \(6\)