Question

Find two numbers such that their sum ,their difference, and sum of their squares are in the ratio 3:1:40.

Answer 1

Let us suppose the suppose the two numbers and x and y.

Now, we have been given that their sum ,their difference, and sum of their squares are in the ratio 3:1:40. Thus, we have

$x+y:x-y:x^2+y^2= 3:1:40$

Therefore, we have

[tex]\frac{x+y}{x-y}=\frac{3}{1}\\ \\ 3x-3y=x+y\\ 2x=4y\\ x=2y...........(1)\\ \text{Also, we have}\\ \frac{x-y}{x^2+y^2}=\frac{1}{40}\\[/tex]

On substituting the value of x from equation 1, we get

[tex]\frac{2y-y}{4y^2+y^2}=\frac{1}{40}\\ \\ \frac{y}{5y^2}=\frac{1}{40}\\ \\ \frac{1}{5y}=\frac{1}{40}\\ \\ 5y=40\\ \\ y=8[/tex]

Hence, from equation 1, the value of x is given by

[tex]x=2\times 8\\ x=16[/tex]

Hence, the required numbers are 16 and 8.