Question
The roots of the given equation (p-q) x^2+(q-r) x+(r-p)=0 are :
BITSAT
Mathematics
Quadratic Equation
2023
BITSAT 2023 (Memory Based Paper 1)
BITSAT Mathematics Question (2023) — Solution
Options
- A. p - q r - p , 1
- B. q - r p - q , 1
- C. r - p p - q , 1
- D. None of these
Answer
C. r - p p - q , 1
Step-by-step solution
Given equation is (p-q) x^2+(q-r) x+(r-p)=0 By using formula for finding the root viz: -b b^2-4 a c 2 a , we get x= (r-q) (q-r)^2-4(r-p)(p-q) 2(p-q) x= (r-q) (q+r-2 p) 2(p-q) = r-p p-q , 1
Related: Mathematics — Quadratic Equation · All PYQ Banks