Question

Can someone help me? I have no idea how to do this? I will Fan and Medal!

Answer 1

Solve
\[\sqrt{6g-23} = \sqrt{12-g}\]

Answer 2

A.) 5
B.) 7
C.) 11
D.) 35

Answer 3

Solve
\[\sqrt{d^2 -11} =5\]

A.) 4,-4
B.) 5.-5
C.) 5,6
D.) 6,-6

Answer 4

@mathmale can you please explain how to do this?

Answer 5

Square both sides of the equation

Answer 6

how would I do that with a variable?

Answer 7

The man told you to square both sides for a reason. Doesn't matter if there are variables in the square root.

Answer 8

If you square both sides, your equation will look like this:

6g-23=12-g

From there, solve for g.

Answer 9

Okay so next itll look like this right
7g-23=12

Answer 10

Well, you did one step. Now isolate the variable.

Answer 11

then
7g= 35

Answer 12

g=5 right?

Answer 13

Yes.

Answer 14

thanks :)

Answer 15

How about the second one? Good job on the first. Medal for two correct answers :-)

Answer 16

\[d^2 -11 = 5\]

Answer 17

\[d^2 =16\]

Answer 18

\[d= 4,-4\]

Answer 19

Plug \(d=4\) into the original equation. Does it work?

Answer 20

ya

Answer 21

No, it doesn't work!

\[\sqrt{d^2-11} = 5\]\[d=4\]\[\sqrt{16-11} = 5\]\[\sqrt{5} = 5\]Not in my universe!

Answer 22

@SnowCrystal come back and fix your work!

Answer 23

I have brought shame to this community. T.T I was being lazy in grading work.

Answer 24

oh i forgot the square root

Answer 25

@SnowCrystal do you have a new, improved answer for me? You can still earn that medal, I'm feeling generous this morning :-)

Answer 26

6,-6

Answer 27

I'm actually a little bit pleased that you made this mistake, because it's a natural introduction to the other point I want to make.

Yes, let's test them out:

\[\sqrt{(6)^2-11} = 5\]\[\sqrt{36-11}=5\]\[\sqrt{25} = 5\checkmark\]\[\sqrt{(-6)^2-11}=5\]\[\sqrt{36-11}=5\checkmark\]

Answer 28

So the other point is that whenever you solve an equation with a radical sign in it by squaring both sides, it is important to test all of your answers. The reason for this is that if you square the radical, you may introduce what is called an extraneous solution, a solution which does not satisfy the original equation!

This usually happens when you have radicals on one side but not the other.

\[\sqrt{2x+4}=x-2\]\[2x+4=(x-2)^2\]\[2x+4=x^2-4x+4\]\[x^2-6x=0\]\[x(x-6)=0\]\[x=0,x=6\]
Looks good so far, right? But let's try them out!

\[\sqrt{2(6)+4} = 6-2\]\[\sqrt{16}=4\]\[4=4\checkmark\]

\[\sqrt{2(0)+4} = 0-2\]\[\sqrt{4} = -2\]Uh oh. If we don't have a - in front of the radical sign, it means take the principal root, which is the positive one here.
\[2 = -2\]is not true, so \(x=0\) is an extraneous solution.

Rather than trying to memorize rules for knowing when an extraneous solution might occur, I find it safer to simply test all of the solutions whenever I square both sides in the process of finding the solution. In fact, I'm a pretty big fan of checking any answer, whenever possible.

Answer 29

Thank you for all ur help :)