NDA2017MathematicsQuadratic EquationActual
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
Correct answer
B. (p-q) /(q-r), 1
Step-by-step solution
Given equation, (q-r) x^2+(r-p) x+(p-q)=0 On observing the equation, it is clear that 1 is 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 . (1)( )= p-q q-r = p-q q-r .