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.
APPEARS IN
संबंधित प्रश्न
The product of two consecutive integers is 56. Find the integers.
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`.
Divide 15 into two parts such that the sum of their reciprocals is `3/10`.
Three positive numbers are in the ratio `1/2 : 1/3 : 1/4`. Find the numbers if the sum of their squares is 244.
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Given that the sum of the squares of the first seven natural numbers is 140, then their mean is ______.
The sum of a number and its reciprocal is 5.2. The number is ______.
The product of two whole numbers, each greater than 4, is 35; the numbers are ______.
