quadratic equation solver- I need help?

Question

you have two numbers. The difference of the two numbers is 12. The product of the two numbers is 17. Multiply the larger of the two numbers by 100, add 50 times the smaller number to that, round the total to the nearest whole number

That was the equation i was given, I was told it is correct and is possible, but unfortunately I am unable to complete it.

Answer 1

So I have x and y being my two numbers.
The difference of the two numbers is 12.
so x-y=12
The product of the two numbers is 17.
that is xy=17
given x>y
x*100 +50*y = some number

First lets solve for the variables, I know that i can rewrite the first equation as x=12+y so with that i can plug it in the other equation to be (12+y)y=17
multiply it through and moving the 17 over gives
12y + y^2 - 17 = 0
now i can use the quadratic formula
y=(-b+/-radical(b^2-4ac))/(2a) ; given ax^2 +bx + c=0
and see that
y= -13.28 and 1.28 (you get two answers)
pluggin those values of y into x=12+y gives me
x= 1.28 for y=-13.28 and x= 13.28 for y = 1.28
first we check which is the right answer by pluggin into the original problem and the condition x>y
so x-y=12 and xy=17
13.28 - (1.28) = 12 checks
1.28 * (13.28) = 16.9 = ~17 (since i rounded the answer for the quadratic )
and we throw away the ones that didn't work (sometimes it gives us answers that don't meet our conditions)

x*100 +50*y =some number
(13.28)*100 +50 (1.28) = 1392
and there you have it (if i did it correctly)



**edit I didnt do it correctly , haha mixed up the signs but i fixed it now.... ALWAYS CHECK YOUR ANSWERS!!!!!!****!!*!*!**!*!*!

Answer 2

I'm wondering. Because you said the product of the numbers is 17. 17 is a prime number, and you will never get 17 as a product of two whole numbers. Please clarify.

Answer 3

17 is a prime number and therefore does not have factors.
Plan B?!!!
Please post amended version of question.