Question

Factoring a square root...

sqr(x-2)-sqr(2x+6)-1

I need to turn this into a quadratic equation.

Answer 1

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

Answer 2

Square everything and solve normally.

Answer 3

ACK! That was so simple... I don't know why I didn't remember that!!!!

Answer 4

Now would the negative signs turn to positives?

Answer 5

I mean the central negative sign and the sign in front of 1

Answer 6

Nope. All of the quantities are squared. Like, each of the roots and it's -(1^2) for the constant.

Answer 7

Okay great, that helped out so much!

Answer 8

So just to be sure, x=-1?

Answer 9

Another way to think of it would be to write out the exponents of each quantity, like the root is to the one half, then if you multiply (1/2)(2), you get (2/2)=1 right? So the exponents cancel but the quantity inside stays the same.

Answer 10

No, it doesn't equal -1, I missed an addition step.

Answer 11

The result would be something like -x+3

Answer 12

x=3

Answer 13

You get -x+3 because if you square everything, you get x-2-x+6-1, so combine like terms.

Answer 14

It seems right until I plut it into
\[\sqrt{2(3)+6}\]

That turns into:\[\sqrt{12}\]
But there is no complete square root of 12, so whatever it is -1, it can't be right.

Answer 15

That leaves me with:

1-\[1-\sqrt{12}-1=0\]

Answer 16

No, that's not it. The result is not a number unless you have something to set it equal to. You get an expression as an answer.

Answer 17

In the original problem, the -1 was =1 and I moved it to the left side, I just didn't include that step.

Answer 18

What was the left side originally?

Answer 19

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

Answer 20

Either way, if I square the 1 on either side, I still end up with -1.

Answer 21

No, I mean the full equation. In your question, you gave an expression. An equation must have an equal sign. Was zero on the other side of the equation originally? Or was nothing there?

Answer 22

The last equation I posted was the full equation.

Answer 23

Then your answer will be an expression, and you cannot move it to the other side.

Answer 24

It says I have to solve for x though, I have to create 0 to solve for x.

Answer 25

Oh so you mean it was given to you already set equal to one? In the problem they gave you?

Answer 26

Yes, exactly. I just left out that step in my original post. The intent was to save you some work, but I will know better next time :P

Answer 27

Lol well you have to say that you need to solve for x, otherwise the answer will be an expression.

Then here's what you do:

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

Which is the same as:

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

In order to solve, we must get rid of the root, and we do this by squaring the entire equation.
\[(x-2)^{(1/2)(2)} - (x+6)^{(1/2)(2)} = 1^{2}. = x - 2 - x + 6 = 1\]

Answer 28

Oh! I tried to do that at first, but I didn't think to square those answers.

Answer 29

Yeah, just do all that, and then solve for x like a normal linear equation.

Answer 30

Thank you, ahhhh, I was so close :P

Answer 31

Haha no worries. And I just realized I forgot the 2x so just pretend it's there lol