Difference of a number and its square is equal to the difference of its square and 1 less than twi

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24 Dec 2010 Find the value of X

Difference of a number and its square is equal to the difference of its square and 1 less than twice
of the number. Product of the values of x is

OPtion

1) 0
2) 3/4
3) 2/3
4) 3/2
5) 1
6) 1/2
7) 4/3
8) 4/9
9) 9/2
10) 4

Solution

x – x2 = x2 – (2x-1)

On solving this equation we get x = 1, 1/2

Correct Option: 6 Best Solution (6)

KALPANA G   13 years AGO

take the number as 'x'.
Difference of a number and its square is equal to the difference of its square and 1 less than twice
of the number.
so, x-x^2 = x^2-(3x-1) => x^2-4x+1 = 0,
then , x = (2+square root(2))/2 (or) (2- square root(2))/2,
so product of the values of x = (2+square root(2))/2*( 2- square root(2))/2 = (4-2)/4= 2/4 = 1/2.

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Sathish   13 years AGO

Let the number be x ,
x - x^2 = x^2 - (2x-1) ,
2x^2-2x+1-x=0 ,
2x^2 - 3x + 1 = 0 ,
2x^2 - x - 2x + 1 = 0 ,
x(2x-1)-1(2x-1)=0
(x-1)(2x-1)=0
x=1,1/2

So the product of thevalues of x is 1 * (1/2) = 1/2

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Arneet   13 years AGO

1*1/2=1/2

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Dr.Dipin Singh   13 years AGO

x-x^2= x^2-(2x-1) or 2x^2-3x+1=0
solving it, we get, x=1,1/2
Then 1*1/2=1/2

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JAYEETA DESHMUKH   13 years AGO

x-x^2= x^2-(2x-1)
or, 2x^2-3x+1=0
or, (x-1)(2x-1)=0
or, x= 1, (1/2)
So, product of the values of x= 1*(1/2)= (1/2)..... Option 6)

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suchandra raavi   13 years AGO

Let the number be x.Difference of x and its square is x-x².Difference of its square and 1 less than twice of the number is x²-(2x-1).Given that both these values are equal. Equate and solve we get 2x²-3x+1=0 => 2x²-2x-x+1=0 => 2x(x-1)-1(x-1)=0 => (2x-1)(x-1)=0 => x=1/2 or 1.Product of 1/2 and 1 is 1/2.

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