Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.
Answers
SOLUTION :
Let the three consecutive integers be x, x +1, x + 2
A.T.Q
x² + (x +1)(x + 2) = 46
x² + x² + 3x + 2 = 46
2x² + 3x + 2 - 46 = 0
2x² + 3x - 44 = 0
2x² - 8x + 11x - 44 = 0
[By middle term splitting]
2x(x - 4) + 11 (x - 4) = 0
(2x + 11) (x - 4) = 0
(2x + 11) = 0 or (x - 4) = 0
2x = - 11 or x = 4
x = - 11/2 or x = 4
Since, x is a positive integer, so x ≠ -11/2
Therefore , x = 4
First positive integer = x = 4
Second positive integer =( x + 1) = 4 + 1 = 5
Third positive integer =( x + 2) = 4 + 2 = 5
Hence, the three consecutive integers are 4, 5 and 6.
HOPE THIS ANSWER WILL HELP YOU...
Answer:
SOLUTION :
Let the three consecutive integers be x, x +1, x + 2
A.T.Q
x² + (x +1)(x + 2) = 46
x² + x² + 3x + 2 = 46
2x² + 3x + 2 - 46 = 0
2x² + 3x - 44 = 0
2x² - 8x + 11x - 44 = 0
[By middle term splitting]
2x(x - 4) + 11 (x - 4) = 0
(2x + 11) (x - 4) = 0
(2x + 11) = 0 or (x - 4) = 0
2x = - 11 or x = 4
x = - 11/2 or x = 4
Since, x is a positive integer, so x ≠ -11/2
Therefore , x = 4
First positive integer = x = 4
Second positive integer =( x + 1) = 4 + 1 = 5
Third positive integer =( x + 2) = 4 + 2 = 5
Hence, the three consecutive integers are 4, 5 and 6.
HOPE THIS ANSWER WILL HELP YOU...