Question

Solve. Check for extraneous solutions.

sqrt2x+6 - sqrtx-1=2

Answer 1

\[\sqrt{2x+6}-\sqrt{x-1}=2\]
@Hero

Answer 2

Alright, any idea how to start?

Answer 3

I have no clue, that's why I need it explained in detail.

Answer 4

Well first lets start off by getting rid of the squareroots on the left side, you can do that by squaring both sides.

Answer 5

\[(\sqrt{2x+6}-\sqrt{x-1})^2 = 2^2\]

What do you get?

Answer 6

I'm not positive, I know it has to be x=

Answer 7

Whale, obviously, you're solving for the only variable C:

What's \(\ (\sqrt{2x+6}-\sqrt{x-1})^2 = 2^2 \) ?

Answer 8

I don't know, I've tried 2, 0, 3, 5, and 10 in place of x

Answer 9

Square both sides is a good start

Answer 10

i don't understand why you're plugging in numbers :o, we have to solve for x.

Answer 11

*Cough* You want to get rid of the square root, so you do the inverse, which is squaring it. Squaring both sides, you get? >.>

Answer 12

I'm so confused

Answer 13

\(\sqrt{2x+6}-\sqrt{x-1}=2\)

First add \(\sqrt{x - 1}\) to both sides

Answer 14

@ChasingMaggie, are you still here?

Answer 15

Yes @Hero

Answer 16

Did you add \(\sqrt{x - 1}\) to both sides?

Answer 17

I really don't think I'm following. Would the equation then be \[\sqrt{2x+6} - \sqrt{x-1}+\sqrt{x-1}=2+\sqrt{x-1}\]

Answer 18

Yes and obviously \(-\sqrt{x - 1} + \sqrt{x - 1} = 0\) so you have

\(\sqrt{2x + 6} = 2 + \sqrt{x - 1}\)

Answer 19

And the first part of the equation would then turn into \[\sqrt{2x+6}=\]

Answer 20

Next square both sides:

\((\sqrt{2x + 6})^2 = (2 + \sqrt{x - 1})^2\)

Answer 21

What will the left side simplify to?

Answer 22

Would it be \[\sqrt{4x+12}\]

Answer 23

Actually, in case you no one told you, when you apply a square to a square root, they will cancel since square and square root are inverses of each other.

In this case, for example, \(\sqrt{x^2} = x\)

Answer 24

Likewise \((\sqrt{2x + 6})^2 = 2x + 6\)

Answer 25

So the left side of the equation simplifies to

\(2x + 6 = (2 + \sqrt{x - 1})^2\)

Answer 26

Meanwhile on the right side, we must expand:
\(2x + 6 = (2 + \sqrt{x - 1})(2 + \sqrt{x - 1})\)

Answer 27

What's next?

Answer 28

On the right side, continue expanding. Your class might call it FOIL. Here we just call it multiplying binomials or finding the product of two binomials.

Answer 29

I never learned that. I tested out of my other math class into this one last week, and I'm expected to know all this.

Answer 30

So you never learned how to multiply something like (x + 2)(x + 3)?

Answer 31

No, I know how to do that.

Answer 32

Are you familiar with the distributive property?

Answer 33

Yes

Answer 34

would it be \[2x+6=(\sqrt{x}+1)^{2}\]

Answer 35

never mind, that's not right

Answer 36

@hero can you just help me through the rest?

Answer 37

It's kind of difficult to help you if you're not familiar with distributive property.

Answer 38

What about 2(x + 5) do you know how to expand that?

Answer 39

By the way distributive property is a(b + c) = ab + ac