Advertisements
Advertisements
प्रश्न
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Advertisements
उत्तर
Let the numbers be x and x + 1.
From the given information,
x2 + (x + 1)2 = 41
2x2 + 2x + 1 – 41 = 0
x2 + x – 20 = 0
(x + 5)(x – 4) = 0
x = –5, 4
But, –5 is not a natural number.
So, x = 4.
Thus, the numbers are 4 and 5.
संबंधित प्रश्न
Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
The sum of a number and its reciprocal is 4.25. Find the number.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is `7/10`.
The denominator of a positive fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
The sum of two natural numbers is 5 and the sum of their reciprocals is `5/6`, the numbers are ______.
Two integers differ by 2 and sum of their squares is 52. The integers are ______.
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 ______.
If 18 is added to a two-digit number, its digits are reversed. If the product of the digits of the number is 24, the number is ______.
The sum of two whole numbers is 18 and their product is 45, the numbers are ______.
