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| Math Central | Quandaries & Queries |
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My teacher didn't explain how to work these out very well. She gave us three problems, i solved two. I have looked elsewhere for help also but I only found problems like this that had to do with multiplication. this is the problem... The sum of two numbers is 9. The sum of the squares of the two numbers is 41. Find the numbers. I can get do this much but then I get stuck... x+y=9 x^2 + y^2=41 -x -x y=9-x then I put this into the equation to the above right but... then when I use the quadratic formula to solve for 'x' I get a negative number inside the radical (square root sign) and you can't get a square from a negative number I appreciate the help. |
Hi Gaby.
You are certainly using the right approach, so I think you must have made an error in your arithmetic somewhere. When I substitute (9-x) in place of y in the equation x2 + y2 = 41, I get a simple quadratic that you can easily factor and get two values of x. Then you can use each of those values to determine two corresponding values of y. I get nice whole numbers and no negative under the radical, so just check your arithmetic.
Stephen La Rocque.>
Hello Gaby,
You first need to translate the question,look at this similar problem;
the sum of two numbers is 7. the sum of the squares of the two nummbers is 25. find the two numbers.
Let x = the first number, Let y=the second number,
then we can create our two equations,
X + Y = 7
X2 + Y2 = 25
If we solve the first equation,
y = 7 - x
and sub it into the other equation,
x2 + (7 - x)2 = 25
We can now multiply this out and solve for x,
x2 + 49 - 14x + x2 = 25
2x2 - 14x + 49 - 25 = 0
2x2 - 14x + 24 = 0 (divide by 2)
x2 - 7x + 12 = 0 (factor)
(x - 4)(x - 3) = 0 so
x = 4 or x = 3
Sub back into original equations to check:
3 + 4 = 7
32 + 42 =25
9 + 16 = 25
So one number is 3 and the other number is 4.
Melanie
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