Advertisements
Advertisements
Question
There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?
Advertisements
Solution
Let the first integer = x
then second integer = x + 1
and third integer = x + 2
Now according to the condition,
x2 + (x + 1)(x + 2) = 154
⇒ x2 + x2 + 3x + 2 - 154 = 0
⇒ 2x2 + 3x - 152 = 0
⇒ 2x2 + 19x - 16x - 152 = 0
⇒ x(2x + 19) - 8(2x + 19) = 0
⇒ (2x + 19)(x - 8) = 0
Either 2x + 19 = 0,
then 2x = -19
⇒ x = `-(19)/(2)`
But it is not possible as it is not an positive integer.
or
x - 8 = 0,
then x = 8
∴ Numbers are 8, (8 + 1) - 9 and (8 + 2) = 10.
APPEARS IN
RELATED QUESTIONS
Solve the following quadratic equations by factorization:
25x(x + 1) = -4
Determine two consecutive multiples of 3, whose product is 270.
For the equation given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of the equation:
(m – 3)x2 – 4x + 1 = 0
For the equation given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of the equation:
x2 – (m + 2)x + (m + 5) = 0
Solve the following quadratic equations by factorization:
`(x-3)/(x+3 )+(x+3)/(x-3)=2 1/2`
Solve the following equation: `"x"^2 - ( sqrt 2 + 1) "x" + sqrt 2 = 0 `
Solve the following equation by factorization
2x2 – 8x – 24 = 0 when x∈I
Find two consecutive integers such that the sum of their squares is 61
Solve the following equation by factorisation :
2x2 + ax – a2= 0
Find the roots of the following quadratic equation by the factorisation method:
`2x^2 + 5/3x - 2 = 0`
