Advertisements
Advertisements
Question
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Solution
Let the two parts be x and y.
From the given information,
= x + y = 20 ⇒ y = 20 – x
= 3x2 = (20 – x) + 10
= 3x2 = 30 – x
= 3x2 + x – 30 = 0
= 3x2 – 9x + 10x – 30 = 0
= x(3x + 10) – 3(3x + 10) = 0
= (3x + 10)(x – 3) = 0
= `x = 3, x = (-10)/3`
Since, x cannot be equal to `(-10)/3`, so, x = 3.
Thus, one part is 3 and the other part is 20 – 3 = 17.
APPEARS IN
RELATED QUESTIONS
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 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.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is `7/10`.
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.
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.
In a two-digit number, the ten’s digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Given that the sum of the squares of the first seven natural numbers is 140, then their mean is ______.
The difference between the digits of a two-digit number is 2 and the product of digits is 24. If tens digit is bigger, the number is ______.
The sum of a number and its reciprocal is 4.25; the number is ______.
