Question

`2x^2+x-4=0`

Answer 1

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.