0

Guest

Courses New

find two natural numbers whose difference is 4 and sum of their squares is 8

find two natural numbers whose difference is 4 and sum of their squares is 8

Grade:8

1 Answers

Santosh Kumar Thakur
20 Points
6 years ago
It is not possible. Because
 
Let the two numbers be A and B where A>B.
Now, A/Q
 A^2 + B^2 =8         ….....(i)
A – B = 4                …......(ii)
 
From eqn. (ii) we get
A = 4 +B
 
Now putting the value of A in eqn (i) we get
     (4+B)^2 + B^2 = 8
=> 16 + B^2 + 8B + B^2 = 8
=> 2B^2 + 8B + 8 = 0
So, B = -2        (By Quadratic Equation)
But  -2 is not a natural number.
Hence It isn.t possible.

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More