`2x^(2) + 3x - alpha - 0 " has roots "-2 and beta " while the equation "x^(2) - 3mx + 2m^(2) = 0 " has both roots positive, where " alpha gt 0 and beta gt 0.`
The roots of the equation
`(q-r)x^(2)+ ( r-p)x + (p-q) = 0 ` are

Options

A. `(r-p)//(q-r),1//2`
B. `(p-q)//(q-r),1`
C. `(q-r) //(p-q),1`
D. `(r-p) //(p-q),1//2`

Answer

B

Explanation

Given equation , `(q-r)x^(2)+(r-p)x+(p-q)=0`
On observing the equation , it is clear that 1 root of equation.
If `x=1, then q-r+r-p+p-q=0`.
` :. 1 " is one root of given equation"`.
Since, the given equation is quadratic equation , we
know that product of roots is `c/a`.
Let the second root be `alpha`
` :. (1) (alpha) = (p-q)/(q-r) rArr alpha = (p-q)/(q-r).`