View Full Version : Need help with maths question...

Spinner

11-18-2007, 09:08 PM

Here it is...

One integer is 5 more than twice another integer. The squares of these two integers have a difference of 312.

Write at least one equation to describe this situation, and use it to find the two integers. Show all your working.

Ugh. I'm just no good at maths... and this is proving a bit of a poser for me:confused:. Any help would be greatly appreciated:happy:

I've got as far as this:

x= 2y + 5

x2 - y2 = 312 (NOTE: The two's are squares)

I think you have to combine the equations or something, but I am unsure how to do it...

pikahero2

11-18-2007, 09:16 PM

Uh...I'd help but I dont understand a word you just said ....:oops:

Im just 11

sonickid01

11-18-2007, 09:16 PM

Here it is...

One integer is 5 more than twice another integer. The squares of these two integers have a difference of 312.

Write at least one equation to describe this situation, and use it to find the two integers. Show all your working.

Ugh. I'm just no good at maths... and this is proving a bit of a poser for me:confused:. Any help would be greatly appreciated:happy:

I've got as far as this:

x= 2y + 5

x2 - y2 = 312 (NOTE: The two's are squares)

I think you have to combine the equations or something, but I am unsure how to do it...

Well, here's what you can do. Use substitution to "combine them."

if x = 2y + 5.

and x^2 - y^2 = 312...

Just replace the x with what it equals. It's that easy and simple. :P

You get (2y+5)^2 - y^2 = 312

Which, if I'm right, becomes:

4y^2 + 25 - y^2

I'll leave you there.

Chromatic Alchemist

11-30-2007, 09:28 PM

What's the square root of 45? There has to be one or the US has had a major messup!:rolleyes:

Linoone

11-30-2007, 09:30 PM

What's the square root of 45? There has to be one or the US has had a major messup!:rolleyes:

I think it is 6.900

Though I am only 10 so I do not know.

Chromatic Alchemist

11-30-2007, 09:35 PM

Nope. That would be something like: 46.??. Besides, it's like a 7th grade problem.

Linoone

11-30-2007, 09:36 PM

Nope. That would be something like: 46.??. Besides, it's like a 7th grade problem.

Oh, well i was close...and I am in 5th grade, so I guess I knew I wouldn't get it.

Chromatic Alchemist

11-30-2007, 09:37 PM

Oh, well i was close...and I am in 5th grade, so I guess I knew I wouldn't get it.

Well, you tried. I'll give you cred for that.

sonickid01

11-30-2007, 11:55 PM

Depending on your grade and how much you know of math, you may or may not get this:

Square root of 45? You could find it in decimal form by checking squares around it, and guestimating in the middle. 36<45<49, so it's pretty close to seven. 6.(something above 5), I guess.

To leave it in square root form, you find squares that are divisors (factors) of 45. We have 9 for one thing, since 9 * 5 = 45. That's really it.

So we divide the 45 by the 9 and arrange it like this: 9*(sqrt5). Then you take the square root of 9 and replace it with it to get 3*(sqrt5).

I think that's right. I'm 75% sure, but meh. It may be the other way around, I'm not too sure. So you'd put 9 in the square root and get 5*(sqrt3).

I think.

Brainiac

12-01-2007, 12:17 AM

Well, here's what you can do. Use substitution to "combine them."

if x = 2y + 5.

and x^2 - y^2 = 312...

Just replace the x with what it equals. It's that easy and simple. :P

You get (2y+5)^2 - y^2 = 312

Which, if I'm right, becomes:

4y^2 + 25 - y^2

I'll leave you there.

This is almost right... You got it real close, but you actually have to distribute the square, like this.

You write down (2y+5)^2 as (2y+5)X(2y+5) and then you distribute it, by doing each one, can't really type it out in a way that illustrates it well, but you end up with 4y^2 + 20y + 25, then you put it back into the problem.

4y^2 + 20y + 25 - y^2 = 312

Then you simplify it a bit, by adding and subtracting like terms, and come out with:

3y^2 + 20y + 25 = 312

then you have to complete the square, something that I'm a little hazy on, but you might have a little more fresh if your taking this class right now...

terry

12-01-2007, 12:21 AM

:eek::eek::eek:wow. your smart. at what grade level is the problem?

Chromatic Alchemist

12-01-2007, 12:24 AM

This is almost right... You got it real close, but you actually have to distribute the square, like this.

You write down (2y+5)^2 as (2y+5)X(2y+5) and then you distribute it, by doing each one, can't really type it out in a way that illustrates it well, but you end up with 4y^2 + 20y + 25, then you put it back into the problem.

4y^2 + 20y + 25 - y^2 = 312

Then you simplify it a bit, by adding and subtracting like terms, and come out with:

3y^2 + 20y + 25 = 312

then you have to complete the square, something that I'm a little hazy on, but you might have a little more fresh if your taking this class right now...

Could you figure out my question.

terry

12-01-2007, 12:27 AM

I just looked at the numbers and went like "WTF?!?!"

Chromatic Alchemist

12-01-2007, 12:28 AM

I just looked at the numbers and went like "WTF?!?!"

LOL!:crackup::crackup:

Brainiac

12-01-2007, 12:55 AM

Could you figure out my question.

well, I don't know, it depends on what question you wanted me to figure out... If you mean the square root of 45, then it is like 6.70820.....bunch of other numbers that I don't want to figure out, but anyways, I think I remembered how to complete the square...

You need to make 3y^2 + 20y + 25 a perfect square number so that you can take the square root of it... so to do that, you should add 8 & 1/3 to your 312 and add it to the other side, so that you have

(3y^2 + 20y + 33 & 1/3) = 320 & 1/3

Yes, the parentheses are around it for a reason... You need to take the square root of both sides, and when you do, you come out with...(I'll use ^ 1/2 for square root...)

(3^1/2)y + 10/(3^1/2) = (303 & 1/3)^1/2

after that, you can just subtract the 10/(3^1/2) from both sides, which would be a lot easier if you use a calculator, but after that, you devide both sides by the square root of three, then you just have y=some number... then you can plug it into the first equation and figure out x. then you should probbably plug both numbers into the second equation, just to make sure that it works out right...

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