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Question 66038This question is from textbook an incremental development
: find three consectuitve intergers such that twice the product of the first and second exceeds the square of the third by 4
This question is from textbook an incremental development
Answer by praseenakos@yahoo.com(507)   (Show Source):
You can put this solution on YOUR website! QUESTION:
find three consectuitve intergers such that twice the product of the first and second exceeds the square of the third by 4
ANSWER:
Here we have three consecutive numbers.....
So we can take three consecutive numbers as (x-1), x , (x+1)
Product of first and second is (x-1)*x = x^2 - x
Twice the Product of first and second is = 2(x^2 - x )
==> = 2x^2 - 2x
Now square of the third is = (x+1)^2
==> x^2 + 2x + 1
according to the question,... twice the product of the first and second exceeds the square of the third by 4
which implies.....
2x^2 - 2x = x^2 + 2x + 1 + 4 ( exceeds 4 means + 4 )
==> 2x^2 - 2x = x^2 + 2x + 5
subtract x^2 from bioth sides of the expression...
==>2x^2 - 2x - x^2 = x^2 + 2x + 5 -x^2
==> x^2 - 2x = 2x + 5
Subtract 2x from both sides....
==> x^2 - 2x - 2x = 2x + 5 - 2x
==> x^2 - 4x = 5
Again subtract 5 from both sides....
==> x^2 - 4x - 5 = 5 - 5
==> x^2 - 4x - 5 = 0
Here we have quadratic equation...Solve this function to get the value of "x"
Here we use the following method ,....
GFind two numbers such that whose sum is -4 and whose product is -5
Such two numbers are -5 and +1
So we can write the above equation as ....
x^2 - 5x + 1x - 5 = 0 ( split the middle term only)
(x^2 - 5x ) + (1x - 5 )= 0
==> x ( x - 5 ) + 1( x - 5) = 0
==> ( x- 5)(x + 1 ) = 0
==> either ( x- 5)= 0 or (x + 1 ) =0
( x- 5)= 0 ==> x = 5 and
(x + 1 )= 0 ==> x = -1
So we have two values for x that is 5 and -1
If x = 5, then numbers are (5-1), 5, (5+1)
That is 4,5 and 6
If we take x = -1, then the numbers are (-1-1), -1, (-1+1)
That is -2, -1 and 0
Hope you understood...
Regards.
praseena.
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