Question

If the difference of the roots of equation `x^2 + 2px + q = 0` and `x^2 + 2qx + p =0` are equal than prove that p+q+1 = 0.(`p!=q`)

Answer 1

`x=(-b+sqrt(b^2-4ac))/(2a)`
`=(-b-sqrt(b^2-4ac))/(2a)`
Difference`=(2sqrt(b^2-4ac))/(2a)`
`=sqrt(b^2-4ac)/a`
`sqrt(4p^2-4q)=sqrt(4q^2-4p)`
`4(p^2-q)=4(q^2-p)`
`p^2-q^2+p-q=0`
`(p-q)(p+q)+(p-q)=0`
`(p-q)(p+q+1)=0`
`p+q+1=0`.