Advertisements
Advertisements
Question
The sum of the squares of two consecutive positive integers is 365. Find the integers.
Solution
Let the required two consecutive positive integers be x and (x+1).
According to the given condition,
`x^2+(x+1)^2=365`
⇒`x^2+x^2+2x+1=365`
⇒`2x^2+2x-364=0`
⇒`x^2+x-182=0`
⇒`x^2+14x-13x-182=0`
⇒`x(x+14)-13(x+14)=0`
⇒`(x+14)(x-13)=0`
⇒`x+14=0 or x-13`
∴`x=13` (x is a positive integers)
When `x=13`
`x+1=13+1=14`
Hence, the required positive integers are 13 and 14.
APPEARS IN
RELATED QUESTIONS
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects
Solve the following quadratic equations by factorization:
`2/2^2-5/x+2=0`
Solve the following quadratic equations by factorization:
a2b2x2 + b2x - a2x - 1 = 0
The sum of two numbers is 48 and their product is 432. Find the numbers?
The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.
Divide 57 into two parts whose product is 680.
Solve the following equation: 3x2 + 25 x + 42 = 0
Solve equation using factorisation method:
x(x – 5) = 24
Solve the equation:
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`.
Solve the following equation by factorization
21x2 – 8x – 4 = 0
