# Thread: need help with this question!

1. ## need 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!

2. ## Re: 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.

3. ## Re: 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?

4. ## Re: need help with this question!

Start with x = y + 3 and substitute into the second equation.

5. ## Re: need help with this question!

so i would combine the x2 + y2 = 120 into the x=y+3 somehow?

6. ## Re: need help with this question!

Originally Posted by Prove It
Let x and y be the two numbers. Then x - y = 3 and $\displaystyle \displaystyle x^2 + y^2 = 120$. Solve the equations simultaneously.
Yes, replace the x in $\displaystyle \displaystyle x^2 + y^2 = 120$ with y + 3. This will give you a quadratic equation to solve.

7. ## Re: 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?

8. ## Re: 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.

9. ## Re: 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?

10. ## Re: 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.