GameGuru1 Posted April 10, 2005 Posted April 10, 2005 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 Quote Share this post Link to post Share on other sites More sharing options...

cold_snipe Posted April 10, 2005 Posted April 10, 2005 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 Quote Share this post Link to post Share on other sites More sharing options...

Overclockersdream Posted April 10, 2005 Posted April 10, 2005 (edited) (X)^2+(X+1)^2=421 lol you edited it while i was correction it Edited April 10, 2005 by Overclockersdream Quote Share this post Link to post Share on other sites More sharing options...

road-runner Posted April 10, 2005 Posted April 10, 2005 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. Quote Share this post Link to post Share on other sites More sharing options...

martymcfly Posted April 10, 2005 Posted April 10, 2005 i believe that their way is a *little* bit easier Quote Share this post Link to post Share on other sites More sharing options...

cold_snipe Posted April 10, 2005 Posted April 10, 2005 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. Quote Share this post Link to post Share on other sites More sharing options...

Dasterdly Posted April 10, 2005 Posted April 10, 2005 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. Quote Share this post Link to post Share on other sites More sharing options...

GameGuru1 Posted April 10, 2005 Posted April 10, 2005 Ok thanks guys Quote Share this post Link to post Share on other sites More sharing options...

crash Posted April 10, 2005 Posted April 10, 2005 (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? Quote Share this post Link to post Share on other sites More sharing options...

Dasterdly Posted April 10, 2005 Posted April 10, 2005 I already solved it Quote Share this post Link to post Share on other sites More sharing options...

GameGuru1 Posted April 11, 2005 Posted April 11, 2005 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? Quote Share this post Link to post Share on other sites More sharing options...

Dasterdly Posted April 11, 2005 Posted April 11, 2005 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. Quote Share this post Link to post Share on other sites More sharing options...

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