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प्रश्न
`2x^2+x-4=0`
उत्तर
The given equation is `2x^2+x-4=0`
Comparing it with `ax^2+bx+c=0` we get
`a = 2,b =1and c = -4 `
∴Discriminant, `D=b^2-4ac=(1)^2-4xx2xx(-4)=1+32=33>0`
So, the given equation has real roots.
Now, `sqrtD=sqrt33`
∴ α=`(-b+sqrt(D))/(2a)=(-1+sqrt(33))/(2xx2)=(-1+sqrt(33))/4`
∴ β=`(-b-sqrt(D))/(2a)=(-1-sqrt(33))/(2xx2)=(-1-sqrt(33))/4`
Hence, `(-1+sqrt(33))/4` and `(-1-sqrt(33)/4)` are the roots of the given equation.
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