Question

1. solve for x. what is the extraneous solution? (sqrt{4x + 5}-45=-50) type a response show your work

Explanation:

Step1: Isolate the square root term

Add 45 to both sides of the equation $\sqrt{4x + 5}-45=-50$ to isolate the square root.
$\sqrt{4x + 5}=-50 + 45$
$\sqrt{4x + 5}=-5$

Step2: Analyze the square root's domain

The square root function $y = \sqrt{a}$ is defined such that $a\geq0$ and the output $y\geq0$ for all real - valued $a$ in its domain. Here, the left - hand side $\sqrt{4x + 5}$ represents a square root, so $\sqrt{4x+5}\geq0$ for all $x$ in its domain. But the right - hand side of the equation $\sqrt{4x + 5}=-5$ is negative.

However, if we proceed to solve algebraically (even though we know from the property of square roots that this will lead to an extraneous solution):
Square both sides of the equation $\sqrt{4x + 5}=-5$:
$(\sqrt{4x + 5})^2=(-5)^2$
$4x+5 = 25$

Step3: Solve for x

Subtract 5 from both sides:
$4x=25 - 5$
$4x=20$
Divide both sides by 4:
$x = 5$

Step4: Check the solution

Substitute $x = 5$ back into the original equation:
Left - hand side: $\sqrt{4(5)+5}-45=\sqrt{20 + 5}-45=\sqrt{25}-45=5 - 45=-40
eq-50$

So, the solution $x = 5$ is extraneous because when we substitute it back into the original equation, it does not satisfy the original equation. Also, the step where we had $\sqrt{4x + 5}=-5$ is invalid because the square root of a real number cannot be negative.

Answer:

The extraneous solution is $x = 5$