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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Let x and y be the two numbers. Then x - y = 3 and . Solve the equations simultaneously.
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?
Start with x = y + 3 and substitute into the second equation.
so i would combine the x2 + y2 = 120 into the x=y+3 somehow?
Originally Posted by Prove It Let x and y be the two numbers. Then x - y = 3 and . Solve the equations simultaneously. Yes, replace the x in with y + 3. This will give you a quadratic equation to solve.
im really sorry i just not to sure. would the next step be:
(y+3)^2 + 3 = 120
y^2 + 9+3 = 120?
Almost, (y + 3)^2 = y^2 + 6y + 9, not y^2 + 9. Fix that mistake, simplify, and solve the quadratic equation.
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?
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.
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