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प्रश्न
`2x^2+5sqrt3x+6=0`
उत्तर
The given equation is `2x^2+5sqrt3x+6=0`
Comparing it with `ax^2+bx+c=0,` we get
`a=2, b=5sqrt3 and c=6`
∴ Discriminant, `D=b^2-4ac=(5sqrt3)^2-4xx2xx6=75-48=27>0`
So, the given equation has real roots.
Now, `sqrtD=sqrt27=3sqrt3`
∴α=`(-b+sqrtD)/(2a)=(-5sqrt(3)+3sqrt(3))/(2xx2)=(-2sqrt3)/4=-sqrt3/2`
`β=(-b+sqrtD)/(2a)=(-5sqrt(3)+3sqrt(3))/(2xx2)=(-8sqrt3)/4=-2sqrt3`
Hence, `-sqrt3/2`and `-2sqrt3` are the roots of the given equation.
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