Advertisements
Advertisements
Question
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence; find the three numbers.
Solution
Let the numbers be x – 1, x and x + 1.
From the given information,
x2 = (x + 1)2 – (x – 1)2 + 60
x2 = x2 + 1 + 2x – x2 – 1 + 2x + 60
x2 = 4x + 60
x2 – 4x – 60 = 0
(x – 10)(x + 6) = 0
x = 10, – 6
Since, x is a natural number, so x = 10.
Thus, the three numbers are 9, 10 and 11.
APPEARS IN
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 positive integers is 208. If the square of the large number is 18 times the smaller. Find the numbers.
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is `7/10`.
Three positive numbers are in the ratio `1/2 : 1/3 : 1/4`. Find the numbers if the sum of their squares is 244.
The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.
The product of two consecutive even whole numbers is 24, the numbers are ______.
The sum of the squares of two consecutive integers is 41. 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 ______.
