Question 66038
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.