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

2 Answers
maganbhai P.
Mar 24, 2018

#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)#

Sahar Mulla ❤
Mar 24, 2018

#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)#

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