Question

The solution to square root of the quantity 2 times x plus 3 minus square root of the quantity x plus 1 equals 1 is x = 3. True or false

Answer 1

Well, plug x=3 into the equation. Does it work?

Answer 2

\[\sqrt{2x+3} -\sqrt{x+1} = 1\]\[\sqrt{2*3+3}-\sqrt{3+1} = 1\]
Does it work?

Now, your instructor may be trying to set a trap for you here.

Answer 3

Yes it does! Can you keep helping me?

Answer 4

Solve square root of 3 times x equals square root of the quantity 4 times x minus 1.

Answer 5

The problem says "The solution is..." which implies that there is only 1 solution. Actually, there's another solution, so if you take the problem strictly at its word, the answer would be False, because that is not the only solution. Try out \(x=-1\):
\[\sqrt{2(-1)+3} - \sqrt{-1+1} = 1\]\[\sqrt{-2+3} - \sqrt{0} = 1\]\[\sqrt{1} = 1\]\[1 = 1\]
It turns out that this equation has two solutions, not just one.

However, if you haven't covered having multiple solutions, it may just be a case of sloppy problem writing.

Answer 6

\[\sqrt{3x} = \sqrt{4x-1}\]Is that the new problem?

Answer 7

What do you get if you square both sides of the equation?

Answer 8

Idk what you mean?

Answer 9

You don't know what I mean by what, precisely? Squaring both sides of the equation?

Answer 10

yes what

Answer 11

does that mean

Answer 12

??

Answer 13

Multiply each side by itself

\[\sqrt{3x}*\sqrt{3x} = \sqrt{4x-1}*\sqrt{4x-1}\]Can you simplify that?

Answer 14

Ugh no

Answer 15

\[\sqrt{9x ^{2}}\] is that it for the first part?

Answer 16

\[\sqrt{u}*\sqrt{u} = \sqrt{u*u} = \sqrt{u^2}\]If \(u>0\), \[\sqrt{u^2} =u\]So assuming that \(x>0\), we can just write \[\sqrt{3x}*\sqrt{3x} = \sqrt{4x-1}*\sqrt{4x-1}\]\[3x = 4x-1\]And I hope you can solve that for the value of \(x\)

Answer 17

is it -2

Answer 18

:-(

Does \(3(-2) = 4(-2) -1\)?

Answer 19

-1

Answer 20

Again, try your solution before asking me if you are correct.

Does \[3(-1) = 4(-1) - 1\]

Answer 21

1

Answer 22

That's better. Now, because we squared a square root while solving, we need to test our solution in the original equation. It may or may not work! If it does not, it is what is called an extraneous solution.

\[\sqrt{3(1)} = \sqrt{4(1) -1}\]Is that true?

Answer 23

yes!

Answer 24

Solve square root of the quantity two x plus one equals x plus one for x.

Answer 25

\[\sqrt{2x+1} = \sqrt{x+1}\]Okay, what's your first step here going to be?

Answer 26

(you ought to be able to give me the answer to this one with a few seconds thought, no algebra needed)

Answer 27

I think it's 0

Answer 28

Yes, that is correct.

Answer 29

Solve five equals square root of the quantity x squared plus sixteen for x.

Answer 30

But how would you go through the motions of solving it?

Answer 31

I think the answer is
3

Answer 32

\[5 = \sqrt{x^2+16}\]

Okay, yes, x = 3 is a solution. Is it the only solution?

Answer 33

\[\pm3\]

Answer 34

Much better.

Answer 35

Solve square root of the quantity k minus seven equals the square root of k, minus one for k.

Answer 36

Okay, it's time for you to learn to use the Equation editor button...

Answer 37

scary

Answer 38

\[\sqrt{k-7} = \sqrt{k}-1\]

Answer 39

How are you going to do this?

Answer 40

I think it's 16

Answer 41

Okay.

Answer 42

Is that the answer?

Answer 43

Does it work?

Answer 44

It doessss

Answer 45

Are there any other possible solutions?

Answer 46

No I don't think so ?
Solve -8 +\[\sqrt{5\times-5}\]=-3

Answer 47

This one I need help on

Answer 48

Look at the graph. Move over to 3 on the x-axis. Move up to the curve. Go to the left until you meet the y-axis. Where are you on the y-axis?