Answer:
Step-by-step explanation:
We are already given the following formulas :-
[tex]\blue{\alpha + \beta = \dfrac{-b}{a}}[/tex]
[tex]\red{\alpha \times \beta = \dfrac{c}{a}}[/tex]
Given that the polynomial is :-
[tex]\sf{6x^2 + 0x - 3}[/tex]
Now, we will use factorisation as our first weapon :-
6x² -3 = 0 [As we have to find the zeroes of the equation]
6x² = 3
x² = [tex]\dfrac{3}{6}[/tex]
x² = [tex]\dfrac{1}{2}[/tex]
x = [tex]\dfrac{1}{\sqrt{2}}[/tex]
Now, we can take one zero as 1/Root 2 and other -1/root 2
[tex]\dfrac{1}{\sqrt{2}} + \dfrac{-1}{\sqrt{2}} | \dfrac{0}{6} = 0[/tex]
0 |0 Hence verified.
[tex]\dfrac{1}{\sqrt{2}} \times \dfrac{-1}{\sqrt{2}} | \dfrac{-3}{6} = 0[/tex]
[tex]\dfrac{1}{2} | \dfrac{1}{2}[/tex]
Hence, proved.
