Find the zeros of #6x^2-3#?

1 Answer
Feb 24, 2018

Answer:

Zeros are #1/sqrt2# and #-1/sqrt2#

Explanation:

Here in #6x^2-3#, there is no middle term and polynomial appears as difference of remaining two terms. In such cases, we can write it as #a^2-b^2# and factors are #(a+b)(a-b)#, while zeros can be obtained by putting #a+b=0# and #a-b=0#.

Here, however, one desires to use the method of splitting the middle term and for this one can add and subtract #ab#. Here as #a^2=6x^2#, our #a=xsqrt6# and as #3=(sqrt3)^2#, we add and subtract #xsqrt6xxsqrt3=sqrt18x# and then

#6x^2-3#

= #6x^2+sqrt18x-sqrt18x-3#

= #sqrt6x(sqrt6x+sqrt3)-sqrt3(sqrt6x+sqrt3)#

= #(sqrt6x-sqrt3)(sqrt6x+sqrt3)#

Here as middle term is #0#, we have split it into #+sqrt18x-sqrt18x#

and zeros are given by #sqrt6x-sqrt3=0# i.e. #x=sqrt3/sqrt6=1/sqrt2#

and #sqrt6x+sqrt3=0# i.e. #x=-sqrt3/sqrt6=-1/sqrt2#