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प्रश्न
`2x^2+6sqrt3x-60=0`
उत्तर
The given equation is `2x^2+6sqrt3x-60=0`
Comparing it with `ax^2+bx+c=0`
`a=2, b=6sqrt3 and c=-60`
∴ Discriminant, `D=b^2-4ac=(6sqrt3)^2-4xx2xx(-60)=180+480=588>0`
So, the given equation has real roots.
Now, `sqrt(D)=sqrt588=14sqrt3`
∴`α =(-b+sqrt(D))/(2a)=(-6sqrt(3)+14sqrt(3))/(2xx2)=(8sqrt(3))/4=2sqrt(3)`
β=`(-b+sqrt(D))/(2a)=(-6sqrt(3)+14sqrt(3))/(2xx2)=(-20sqrt(3))/4=-5sqrt(3)`
Hence, `2sqrt(3)` and `-5sqrt(3)` are the root of the given equation.
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