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BITSAT2023MathematicsQuadratic EquationActual

The roots of the given equation (p-q) x^2+(q-r) x+(r-p)=0 are :

Options

  1. Ap - q r - p , 1
  2. Bq - r p - q , 1
  3. Cr - p p - q , 1
  4. DNone of these

Correct 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

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