quadratic equation: sum is 20; sum of squares is 202

[Motherof8]

The sum of 2 numbers is 20 and the sum of their squares is 202. Find the numbers. I can't seem to find any numbers that add up to 20 whose squared add up to 202. I've tried all the numbers but they all come out as more than 202. I've even tied some negative numbers and whole numbers with fractions. What am I missing?

[stapel_eliz]

The sum of 2 numbers is 20 and the sum of their squares is 202. Find the numbers. I can't seem to find any numbers that add up to 20 whose squared add up to 202. I've tried all the numbers but they all come out as more than 202. I've even tied some negative numbers and whole numbers with fractions. What am I missing?

What was your set-up? What variables did you pick for the two numbers? What equations did you create? Where did this lead?

Please be complete. Thank you!

[Motherof8]

I guess the equation would be x +y -20=0 and x squared + Y squared -202 = 0. I didn't know exactly how to solve an equation with regular variables and those that are squared, so I just plugged in certain values for the variables to see if they worked. 10 + 10 =20. 100 + 100 = 200, not 202. 4x 4 = 16, and 16 x 16 is 256, which is more than 202.13 squared is 169 and 7 squared is 49, which adds up to 218 which is more than 202. 12 squared is 144, and 8 squared is 64, which still add up to 208, more than 202. Negative numbers squared are positive, so that wouldn't work.

[Luke53]

I guess the equation would be x +y -20=0 and x squared + Y squared -202 = 0. I didn't know exactly how to solve an equation with regular variables and those that are squared, so I just plugged in certain values for the variables to see if they worked. 10 + 10 =20. 100 + 100 = 200, not 202. 4x 4 = 16, and 16 x 16 is 256, which is more than 202.13 squared is 169 and 7 squared is 49, which adds up to 218 which is more than 202. 12 squared is 144, and 8 squared is 64, which still add up to 208, more than 202. Negative numbers squared are positive, so that wouldn't work.
Motherof8

How about 9 and 11?
Luke.

[stapel_eliz]

I didn't know exactly how to solve an equation with regular variables and those that are squared...

Are you saying that you haven't studied how to solve any equations yet, or only that you need to learn how to solve quadratic equations (that is, the equations with squared variables)?

[Motherof8]

I was wondering how to solve an equation where there is an ordinary x and y and the other one is squared. If one divides the variables, one comes up with x and y. Wduld one divide 20 into 202 to complete the problem?

Anyway, I tried 9 and 11. It worked so I was shortsighted.

[stapel_eliz]

If one divides the variables, one comes up with x and y.

I'm sorry, but I don't understand what you mean by this...?

Since one of the equations is "x + y = 20", you can solve for "y = 20 - x". Then plug this into the other equation in place of "y", and end up with a regular quadratic equation, which you can solve by the usual methods.

[Motherof8]

I substituted 20 - x for y. I came up with (20 - x ) squared + x squared - 202 = 0. 20 -x squared is 400 -40x + 2x squared. 400 -202 =198. The common factor in 198, 40, and 2x squared is 2. I came up with this equation: 2(x squared - 20 + 99). How can I solve it from there?)

[stapel_eliz]

20 -x squared is 400 -40x + 2x squared.

No; the x, when squared, is just x², not 2x².

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Move the 198 to the left-hand side, and divide through by 2:

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Now list the factor pairs of 99: 1 and 99, 3 and 33, 9 and 11. Which pair adds to twenty?