Question

Find the roots of the following quadratic equation: 

`x^2-3sqrt5x+10=0`

Answer 1

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`.