Advertisements
Advertisements
प्रश्न
There are three consecutive integers such that the square of the first increased by the product of the first increased by the product of the others the two gives 154. What are the integers?
उत्तर
Let three consecutive integer be x , (x + 1) and (x + 2)
Then according to question
x2 + (x + 1)(x + 2) = 154
x2 + x2 + 3x + 2 = 154
2x2 - 3x + 2 - 154 = 0
2x2 + 3x - 152 = 0
2x2 - 16x - 19x - 152 = 0
2x(x - 8) + 19(x - 8) = 0
(x - 8)(2x + 19) = 0
(x - 8) = 0
x = 8
Or
2x + 19 = 0
2x = -19
x = -19/2
Since, x being a positive number, so x cannot be negative.
Therefore,
Whenx = 8then other positive integer
x + 1 = 8 + 1 = 9
And
x + 2 = 8 + 2 = 10
Thus, three consecutive positive integer be 8, 9, 10.
APPEARS IN
संबंधित प्रश्न
Solve for x
`(2x)/(x-3)+1/(2x+3)+(3x+9)/((x-3)(2x+3)) = 0, x!=3,`
Solve the following quadratic equations by factorization:
3x2 − 14x − 5 = 0
Sum of two numbers is 16. The sum of their reciprocals is 1/3. Find the numbers.
The sum of ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man's age at the time. Find their present ages.
Solve : x2 – 11x – 12 =0; when x ∈ N
Solve each of the following equations by factorization:
`2x^2-1/2x=0`
The sum of the squares to two consecutive positive odd numbers is 514. Find the numbers.
The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.
Solve the following equation: `(2"x")/("x" - 4) + (2"x" - 5)/("x" - 3) = 25/3`
Solve the following equation by factorisation :
`sqrt(x + 15) = x + 3`