Question

Obtain a quadratic polynomial with following conditions?? 1. the sum of zeroes=1/3,the product of zeroes=1/2

Answer 1

`6x^2-2x+3=0`

Explanation:

The quadratic formula is `x=(-b+-sqrt(b^2-4ac))/(2a)`

Sum of two roots:
`(-b+sqrt(b^2-4ac))/(2a)+(-b-sqrt(b^2-4ac))/(2a)=-(2b)/(2a)=-b/a`

`-b/a=1/3`
`b=-a/3`

Product of two roots:
`(-b+sqrt(b^2-4ac))/(2a)(-b-sqrt(b^2-4ac))/(2a)=((-b+sqrt(b^2-4ac))(-b-sqrt(b^2-4ac)))/(4a^2)=(b^2-b^2+4ac)/(4a^2)=c/a`

`c/a=1/2`
`c=a/2`

We have `ax^2+bx+c=0`

`6x^2-2x+3=0`

Proof:
`6x^2-2x+3=0`

`x=(2-sqrt((-2)^2-4(6*3)))/(2*6)=(2+-sqrt(4-72))/12=(2+-2sqrt(17)i)/12=(1+-sqrt(17)i)/6`

`(1+sqrt(17)i)/6+(1-sqrt(17)i)/6=2/6=1/3`

`(1+sqrt(17)i)/6*(1-sqrt(17)i)/6=(1+17)/36=18/36=1/2`

Answer 2

`6x^2 - 2x + 3 = 0`

Explanation:

If we have a general quadratic equation:

`ax^2 + bx + c = 0 iff x^2 + b/ax + c/a = 0`

And we denote the root of the equation by `alpha` and `beta`, then, we also have:

`(x-alpha)(x-beta) = 0 iff x^2 - (alpha+beta)x + alpha beta = 0`

Which gives us the well studied properties:

# {: ("sum of roots", = alpha+beta, = -b/a), ("product of roots", = alpha beta, = c/a) :} #

Thus we have:

# {: ( alpha+beta, = -b/a, = 1/3), ( alpha beta, = c/a, =1/2) :} #

So the sought equation is:

`x^2 - "(sum of roots)"x + "(product of roots)" = 0`

i.e.:

`x^2 - 1/3x + 1/2 = 0`

And (optionally), to remove the fractional coefficients, we multiply by `6` giving:

`6x^2 - 2x + 3 = 0`