Math Problem

[GameGuru1]

I apologize in advance if this does not belong here...but I can't seem to solve this. Well, I can get an answer by trial and error, but I need a solvable formula.

 

So here it is:

 

"When two consecutive integers are squared and the squares added their sum is 421. What are the possible numbers?"

 

This seems so simple and yet I cannot come up with a good formula...

 

Thx in advance

[cold_snipe]

x^2+(x+1)^2=421

 

EDIT: sorry misread the thing, im rewriting it now

*rewriting*

 

 

ok i re-wrote it i think its good now

 

 

[Overclockersdream]

(X)^2+(X+1)^2=421

 

lol you edited it while i was correction it

Edited April 10, 2005 by Overclockersdream

[road-runner]

I apologize in advance if this does not belong here...but I can't seem to solve this.  Well, I can get an answer by trial and error, but I need a solvable formula.

 

So here it is:

 

"When two consecutive integers are squared and the squares added their sum is 421.  What are the possible numbers?"

 

This seems so simple and yet I cannot come up with a good formula...

 

Thx in advance

460096[/snapback]

 

The two consecutive integers would be two separate variables (x)^2+)(y)^2=421. Try isolating one of the variables and plug in "0" for x or y.

(y)^2=421

 

Y=square root of 421

do the same for x.

 

unfortunately the only way to do this specific equation are by plugging in the answers you recieve from isolating x and y and adding them together and seeing if they equal 421. I hope you know how to add rational numbers with a square root value. If this wasn't in the lesson plan, or not yet discussed, you only need to add the numbers in decimal form by solving the square root value of the equation vis a vis calculator. Let me know, if you have any more questions. I'll solve the equation for you and give you the possible solutions if you like.

[martymcfly]

i believe that their way is a *little* bit easier

[cold_snipe]

The two consecutive integers would be two separate variables (x)^2+)(y)^2=421.  Try isolating one of the variables and plug in "0"  for x or y. 

(y)^2=421

 

Y=square root of 421

do the same for x.

 

unfortunately the only way to do this specific equation are by plugging in the answers you recieve from isolating x and y and adding them together and seeing if they equal 421.  I hope you know how to add rational numbers with a square root value.  If this wasn't in the lesson plan, or not yet discussed, you only need to add the numbers in decimal form by solving the square root value of the equation vis a vis calculator.  Let me know, if you have any more questions.  I'll solve the equation for you and give you the possible solutions if you like.

460106[/snapback]

 

This one is right too, but you know that they are consecutive so y=x+1, you just substitute y for x+1 and you can solve that easily.

[Dasterdly]

x^2 + (x+1)^2 = 421

x^2 + (x^2 +2 x +1) = 421

2x^2 + 2x - 420 = 0

2(x^2 + x - 210) = 0

2(x+15)(x-14) = 0

x = -15, 14

answer = -15, -14 or 14, 15

 

 

edit : you guys complicate things, this is alot easier.

[GameGuru1]

Ok thanks guys

[crash]

(2x + 1)^2 + (2a +1) = 421

 

(2x + 1)^2 = 421 - (2x + 1)

 

4x^2 + 4x + 1 = 420 - 2x

 

4x^2 + 6x = 419

 

Did I get the first few steps right? It's been ten years since my last math class so I'm forgetting most of my math.

 

How do you come up with the solution?

[Dasterdly]

I already solved it

[GameGuru1]

New question.

 

A right angled trianlge has a height 8 cm more than twice the length of the base. If the area of the triangle is 96 cm square, find the dimensions of the triangle.

 

So far I have 96cm^2 = (1/2 x)(2x+8)

 

But solving that I get x=-12 or x=8

 

How do I apply this to the dimensions of the triangle, or am I on the wrong track?

[Dasterdly]

You're on the right track. Throw away the -12 answer, 'cause you can't use it. 8 is the length of the base, 2(8)+8 = 24. then to find the hypotnuse, 8^2 + 24^2 = c^2. The hypotnuse comes out to about 25.3.