If 2 positive numbers differ by 3 and the sum of the squares of these numbers is 120.Find the numbers.?

Any help would be great!

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- Mar 25th 2013, 09:13 PMmathkid182need help with this question!
If 2 positive numbers differ by 3 and the sum of the squares of these numbers is 120.Find the numbers.?

Any help would be great! - Mar 25th 2013, 09:21 PMProve ItRe: need help with this question!
Let x and y be the two numbers. Then x - y = 3 and $\displaystyle \displaystyle x^2 + y^2 = 120$. Solve the equations simultaneously.

- Mar 25th 2013, 11:54 PMmathkid182Re: need help with this question!
hi thanks for the reply. the part you posted is actually the part im stuck on im sure of the next step to solve the equations?

- Mar 25th 2013, 11:55 PMProve ItRe: need help with this question!
Start with x = y + 3 and substitute into the second equation.

- Mar 25th 2013, 11:56 PMmathkid182Re: need help with this question!
so i would combine the x2 + y2 = 120 into the x=y+3 somehow?

- Mar 25th 2013, 11:58 PMProve ItRe: need help with this question!
- Mar 26th 2013, 12:02 AMmathkid182Re: need help with this question!
im really sorry i just not to sure. would the next step be:

(y+3)^2 + 3 = 120

then

y^2 + 9+3 = 120? - Mar 26th 2013, 12:56 AMProve ItRe: need help with this question!
Almost, (y + 3)^2 = y^2 + 6y + 9, not y^2 + 9. Fix that mistake, simplify, and solve the quadratic equation.

- Mar 26th 2013, 01:26 AMmathkid182Re: need help with this question!
so it will be

y2 + 6y + 9 = 120 (take 120 from both side)

= y2 + 6y -117 = 0

and then what I cant fin number that will simply this? - Mar 26th 2013, 04:00 AMHallsofIvyRe: need help with this question!
Sigh. Your left side is (y+ 3)^2. You forgot the other "y^2". Do you know the "quadratic formula" or how to "complete the square"?

You've also done the arithmetic wrong: 9- 120 is NOT equal to -117.