Question

P and Q are the roots of 3x*2-12x+6. Find 1/p*2 - 1/q*2 ?

Answer 1

`1/p^2-1/q^2=2sqrt2`.....`( p < q )`.
Hint: `(x-y)^2=x^2+y^2-2xy=x^2+2xy+y^2-4xy`
`=>(x-y)^2=(x+y)^2-4xy`
please use '^' instead of ' * ' . `i.e.x^2 to`x^2 and not x*2

Explanation:

I think your quadratic equation is

`3x^2-12x+6=0`.

Comparing with `ax^2+bx+c=0`,we get

`a=3, b=-12 and c=6`

If the roots of this equn. are `p and q`, then

`p+q=-b/a and pq=c/a`

`i.e.p+q=-(-12)/3=4 and pq=6/3=2`
Now,
`1/p^2-1/q^2=(q^2-p^2)/(p^2q^2)=((q+p)(q-p))/(pq)^2`,....`( p < q)`

`=>1/p^2-1/q^2=((4)sqrt((q-p)^2))/2^2=sqrt((q-p)^2`

`=>1/p^2-1/q^2=sqrt((q+p)^2-4pq)=sqrt(4^2-4(2)`

`=>1/p^2-1/q^2=sqrt(16-8)=sqrt8=2sqrt2`....`( p < q)`

Answer 2

`3x^2-12x+6=0`
`=>x^2 - 4x +2=0`

Roots, `x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(4+-sqrt(16-4*1*2))/(2)`

`x=(4+-sqrt(8))/(2) = (4+-2sqrt(2))/(2)`

`x=(2+-2sqrt(2))`

To find, `1/p^2 - 1/q^2`

`=>(1/p+1/q)(1/p-1/q)`

`=>(1/(2+2sqrt(2))+1/(2-2sqrt(2)))(1/(2+2sqrt(2))-1/(2-2sqrt(2)))`

`=>(((2-2sqrt(2))+(2+2sqrt(2)))/((2-2sqrt(2))(2+2sqrt(2))))(((2-2sqrt(2))-(2+2sqrt(2)))/((2-2sqrt(2))(2+2sqrt(2))))`

`=>(((2+2))/((2-2sqrt(2))(2+2sqrt(2))))(((-2sqrt(2)-2sqrt(2)))/((2-2sqrt(2))(2+2sqrt(2))))`

`=>((4(-4sqrt2))/((4-8))^2)`

`=>((4(-4sqrt2))/(-4)^2)`

`=>(-sqrt2)`