Advertisements
Advertisements
Question
Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Advertisements
Solution
Let the two numbers be x and x + 5.
From the given information,
x2 + (x + 5)2 = 97
2x2 + 10x + 25 – 97 = 0
2x2 + 10x – 72 = 0
x2 + 5x – 36 = 0
(x + 9)(x – 4) = 0
x = –9 or 4
Since, –9 is not a natural number.
So, x = 4.
Thus, the numbers are 4 and 9.
RELATED QUESTIONS
A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number.
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Divide 15 into two parts such that the sum of their reciprocals is `3/10`.
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
The product of the digits of a two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.
Given that the sum of the squares of the first seven natural numbers is 140, then their mean is ______.
In a school, a class has 40 students out of which x are girls. If the product of the number of girls and number of boys in the class is 375; the number of boys in the class is ______.
The product of two whole numbers, each greater than 4, is 35; the numbers are ______.
The sum of two whole numbers is 18 and their product is 45, the numbers are ______.
