If the difference of two numbers is 9 and their product is 13, what is the sum of their squares?

Question

Two numbers are such that their difference, their sum and their product are in the ratio 1:5:24.
What is the product of these two numbers?

Answer 1

x-y=9
x*y=13
x² + y² = ( x - y )² + 2xy = 9² + 2(13) = 81 + 26 = 107

Comments

You are great

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let the numbers be x and y
we have x-y=9 and x*y=13

we know that (x-y)^2=x^2 + y^2 -2*x*y
so,
81=x^2 + y^2 -2*13
x^2 + y^2= 81+26
x^2 + y^2 = 107
hence,
sum of squares=107

Answer 2

let the numbers be x and y
we have x-y=9 and x*y=13
we know that (x-y)^2=x^2 + y^2 -2*x*y
so,
81=x^2 + y^2 -2*13
x^2 + y^2= 81+26
x^2 + y^2 = 107
hence,
sum of squares=107

Answer 3

x-y=9
xy=13
(x-y)^2=x^2 +y^2-2xy
81=x^2+y^2-26
107=x^2+y^2

Answer 4

(x-y)^2 = x^2 - 2xy + y^2 = 81.
x^2 + y ^2 = 81+ 2xy
2 xy = 2*13 = 26
So, the sum of their squares, x^2 + y^2 = 81+26 = 107.

Answer 5

The numbers be x, and x+9
X(X+9) = 13
X2 + 9x =13 --------------------------(1)

Sum of the squares = x2 + (x+9)2
= x 2+ x2 +18x + 81
= x2 + x2 + 9x + 9x + 81
= x2 + 9x + x2 + 9x + 81
Putting x2 + 9X = 13 ----------------------˃ Answer = 107