Question

One positive integer is 3 less than twice another. The sum of their squares is 698. Find the integers.

Answer 1

Answer:

The value of positive integers are 21.22 and 12.11

Step-by-step explanation:

Given as :

The sum of squares of two integer = 698

Let The one positive integer be x

And The other positive integer be y

According to question

one positive integer = 3 less than twice the other positive integer

So, x = 2 × y - 3

I.e x = 2 y - 3

And x²  +  y² = 698

So, Put the value of x

I.e  ( 2 y - 3 )² +  y² = 698

or, 4 y² + 9 - 12 y +  y² = 698

Or, 5 y² - 3 y - 698 = 0

Now solving this quadratic equation

y = $\frac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$

Or, y = $\frac{3\pm \sqrt{-3^{2}-4\times 5\times -698}}{2\times 5}$

Or, y = $\frac{3\pm \sqrt{13969}}{10}$

Or, y = $\frac{3\pm 118.19}{10}$

∴ y = 12.11 , - 11.51

So , The value of y = 12.11

And the value of x = 2 × 12.11 - 3

I.e x = 21.22

Hence The value of positive integers are 21.22 and 12.11 Answer