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Question
Find the roots of the following quadratic equation:
`x^2-3sqrt5x+10=0`
Solution
The given quadratic equation is`x^2-3sqrt5x+10=0`
On comparing with the standard form of quadratic equation i.e., ax2 + bx + c = 0, we obtain
`a=1,b=-3sqrt5,c=10`
`sqrtD=sqrt(b^2-4ac)=sqrt((-3sqrt5)^2-4xx1xx10)`
`=sqrt(45-40)`
`=sqrt5`
`∴ x=(-b+-sqrtD)/(2a)`
`=(3sqrt5+-sqrt5)/(2xx1)`
`=4sqrt5/2`or `(2sqrt5)/2`
`=2sqrt5` or `sqrt5`
Therefore, the roots of the given quadratic equation are`2sqrt5`and `sqrt5`.
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