The roots of the given equation $(p - q){x^2} + (q - r)x + (r - p) = 0$ are
- $\frac{{p - q}}{{r - q}},1$
- $\frac{{q - r}}{{p - q}},1$
- $\frac{{r - p}}{{p - q}},1$
-
none
For a quadratic equation ax^2 + bx + c = 0, if the sum of coefficients a+b+c = 0, then 1 is a root. Here, (p-q) + (q-r) + (r-p) = 0. Thus, 1 is a root. The product of roots is c/a = (r-p)/(p-q). The other root is (r-p)/(p-q).
Using the method of inspection, we can see that substituting x equals 1 into the equation (p minus q)x squared plus (q minus r)x plus (r minus p) equals 0 yields (p minus q) plus (q minus r) plus (r minus p), which sums to 0, confirming one root is 1. Because the sum of all coefficients is zero, 1 is indeed a root, and for the product of the roots given by c divided by a, we have (r minus p) divided by (p minus q) as the first root. Therefore, the roots are (r minus p) divided by (p minus q) and 1.