Test your algebra skills!

[sandmoth12]

I realize it's against the rules to post homework so I will post the question and the answer as I have already did all the work. You just have to show the work.

square root(2y-3)...subtract...square root(3y+3)...add...square root(3y-2)=0

The answer is y=2.

EDIT: This one was a little tough. Took a few tries to get it right. It's definitely one of the longer problems so attention to detail when playing with signs helps.

---------- Post added 2012-01-29 at 03:55 PM ----------

No one? Am I the only lonely math nerd on this site =(

[Meteoric]

You should end up with 2 possible answers to this one (unless something about the problem rules out a negative solution). I will post my solution briefly, but I want to give you a chance to work on that first.

Edit: Oh, of course, the starting equation would be non-real if y is negative. Silly me.

Here's my solution, then:
Hm... let's see. This might not be the most elegant way to solve the problem, but here's how I'd solve it:
sqrt(2y-3)-sqrt(3y+3)+sqrt(3y-2)=0
Add sqrt(3y+3) to both sides:
sqrt(2y-3)+sqrt(3y-2)=sqrt(3y+3)
Square both sides:
2y-3+2sqrt((2y-3)(3y-2))+3y-2=3y+3
Combine like terms:
5y-5+2sqrt(6y^2-13y+6)=3y+3
Subtract 5y and add 5 to both sides:
2sqrt(6y^2-13y+6)=-2y+8
Divide by 2:
sqrt(6y^2-13y+6)=-y+4
Square both sides:
6y^2-13y+6=y^2-8y+16
Subtract the right side:
5y^2-5y-10=0
You could solve this trinomial directly without too much trouble, but I'm going to divide by 5 first to make it look a little nicer:
y^2-y-2=0
Factors into:
(y-2)(y+1)=0
So y=2 or y=-1. If you're not dealing with imaginary/complex numbers, we can throw out the y=-1 because that would have given us some square roots of negatives in the first equation. So yes, y=2.

[Totle]

So....

Root(2y-3)-Root(3y+3)+Root(3y-2) = 0
Square both sides
(2y-3)-(3y+3)+(3y-2) = 0
2y-3-3y-3+3y-2=0
combine terms
2y-8=0
2y=8
y=4

What did I do wrong?

[Meteoric]

What did I do wrong?

(a+b+c)^2=/=a^2+b^2+c^2
You can't just distribute the squared to each term when there's addition/subtraction in the way, you'd have to multiply them all out and you'd get mixed terms (like FOIL, but with 3 things in each parenthesis instead of 2). That's why I moved one of the square roots to the other side first - to avoid getting so many mixed terms when squaring both sides.

[sandmoth12]

You should end up with 2 possible answers to this one (unless something about the problem rules out a negative solution). I will post my solution briefly, but I want to give you a chance to work on that first.

Edit: Oh, of course, the starting equation would be non-real if y is negative. Silly me.

Here's my solution, then:
Hm... let's see. This might not be the most elegant way to solve the problem, but here's how I'd solve it:
sqrt(2y-3)-sqrt(3y+3)+sqrt(3y-2)=0
Add sqrt(3y+3) to both sides:
sqrt(2y-3)+sqrt(3y-2)=sqrt(3y+3)
Square both sides:
2y-3+2sqrt((2y-3)(3y-2))+3y-2=3y+3
Combine like terms:
5y-5+2sqrt(6y^2-13y+6)=3y+3
Subtract 5y and add 5 to both sides:
2sqrt(6y^2-13y+6)=-2y+8
Divide by 2:
sqrt(6y^2-13y+6)=-y+4
Square both sides:
6y^2-13y+6=y^2-8y+16
Subtract the right side:
5y^2-5y-10=0
You could solve this trinomial directly without too much trouble, but I'm going to divide by 5 first to make it look a little nicer:
y^2-y-2=0
Factors into:
(y-2)(y+1)=0
So y=2 or y=-1. If you're not dealing with imaginary/complex numbers, we can throw out the y=-1 because that would have given us some square roots of negatives in the first equation. So yes, y=2.

Awesome man nearly the exact same way I did it. I don't have any candy but instead /salute =)
I plugged -1 back into the equation and it didn't checkout so I just dropped it.

---------- Post added 2012-01-29 at 04:10 PM ----------

So....

Root(2y-3)-Root(3y+3)+Root(3y-2) = 0
Square both sides
(2y-3)-(3y+3)+(3y-2) = 0
2y-3-3y-3+3y-2=0
combine terms
2y-8=0
2y=8
y=4

What did I do wrong?

Looks like you didn't isolate a single square root on one side of the equal sign before you squared everything.

[kailtas]

I always get sad when i see math on these forums... we use different words in Norway -_-

[Beace]

Yeah, solved it pretty much same way as Meteoric did. Took a few tries to reach the correct answer, and even had to look up some of the rules like:

sqrt(a) x sqrt(b) = sqrt(ab)

Been a long time since I solved these kind of problems.

[lopk]

http://www.wolframalpha.com/input/?i...83y-2%29+%3D+0

This is how I did it.

[Mormolyce]

I always get sad when i see math on these forums... we use different words in Norway -_-

So much for maths being the universal language :P

[Totle]

Oh balls, it's been over a decade since algebra. Hats off to you guys who got it.

[Nawy]

enter this in solve function on my casio calc. +10 internnets?

[Deleted]

I promise I didn't look at the answers above me

sqrt(2y-3) - sqrt(3y+3) + sqrt(3y-2) = 0

sqrt(2y-3) + sqrt(3y-2) = sqrt(3y+3)

2y-3 + 2*sqrt(6y² -13y +6) + 3y-2 = 3y+3

2*sqrt(6y² -13y +6) = -2y + 8

sqrt(6y² -13y +6) = -y + 4

6y² - 13y + 6 = y² - 8y + 16

5y² - 5y - 10 = 0

y² - y - 2 = 0

(y - 2)(y + 1) = 0

y = 2 V y = -1

No real solution for y = -1, so y = 2


I messed up the first time because I made a typo and typed a + instead of a -...