Question

help please work attached

Answer 1

answer choices:
all real numbers
no solution
18
-9
2
15

Answer 2

Just transcribing for the others:

\[\begin{matrix}
\sqrt{4x+1}=0&&&4-(7-y)^{1/2}=0&&&1+\sqrt{x+2}=0\\ \\ \\
\sqrt{z-6}-3=0&&&\frac{1}{6}(12a)^{1/3}=1
\end{matrix}\]

Answer 3

okay so what do i do

Answer 4

Something to keep in mind for the equations in the middle column:
\[x^{1/n}=\sqrt[n]x\]

Starting with the first one:
\[\sqrt{4x+1}=0\]
Squaring both sides:
\[\left(\sqrt{4x+1}\right)^2=0^2\\
4x+1=0\]
What do you get when solving for \(x\)?

Answer 5

x=1/2

Answer 6

Oh sorry, I meant \(\sqrt{4x+1}=3\)...

Close, you have the reciprocal of the actual answer.
\[4x+1=3^2\\
4x=8\\x=2\]

Answer 7

oh so number 1 is 2

Answer 8

Yep.

Next equation:
\[4-(7-y)^{1/2}=0\\
4=(7-y)^{1/2}\]
Raising both sides to power of 2 (squaring both sides):
\[4^2=\left((7-y)^{1/2}\right)^2\\
16=7-y\\
~~~~~~~\vdots\]

Answer 9

y=11

Answer 10

No, that's not it.
\[16=7-y\\
y=7-16\\
y=-9\]
Keep in mind that you have to match answers with the provided list of solutions. If what you get isn't listed, you probably made a slight mistake.

I'll go ahead and post the first steps for the others:
\[1+\sqrt{x+2}=0\\
\sqrt{x+2}=-1\]
What do you know about square roots? (i.e. what can you never get when taking a square root?)

\[\sqrt{z-6}-3=0\\
\sqrt{z-6}=3\\
\left(\sqrt{z-6}\right)^2=3^2\\
~~~~~~~~\vdots \]
\[\frac{1}{6}(12a)^{1/3}=1\\
6\cdot\frac{1}{6}(12a)^{1/3}=6\cdot1\\
(12a)^{1/3}=6\\
\left((12a)^{1/3}\right)^3=6^3\\
~~~~~~~~\vdots\]

Answer 11

umm idk

Answer 12

When you take a square root, you can't get a negative. This means there is no solution to the third equation.

Answer 13

ohhhh yeah sorry i was doing something for my momma