Answer:
Step-by-step explanation:
if a,b are the roots of a quadratic eq. with coefficient of leading term is 1 ,then
(x-a)(x-b)=1
here roots are [tex]Here~roots~are~\frac{\sqrt{3} -1}{2}} ~and~\frac{\sqrt{3}+1 }{2} \\reqd. ~eq.~is~[x-\frac{\sqrt{3} -1}{2} ][x-\frac{\sqrt{3}+1}{2} ]=0\\or~[(x-\frac{\sqrt{3}}{2} })+\frac{1}{2} ][(x-\frac{\sqrt{3}}{2}) -\frac{1}{2}]==0\\or~(x-\frac{\sqrt{3}}{2} )^2-(\frac{1}{2} )^2=0\\or~x^2-\sqrt{3} x+\frac{3}{4}-\frac{1}{4}=0\\or~ x^2-\sqrt{3}x+\frac{1}{2} =0[/tex]
