One of the roots of the equation $\left(q-r\right)x^2+\left(r-p\right)x+\left(p-q\right)=0$ is :
- $\dfrac{r-p}{q-r}$
- $\dfrac{p-q}{q-r}$
- $\dfrac{q-r}{p-q}$
- $\dfrac{r-p}{p-q}$
For a quadratic equation ax^2 + bx + c = 0, if the sum of coefficients (a + b + c) is 0, then x = 1 is a root. Here, (q-r) + (r-p) + (p-q) = 0. Thus, x = 1 is a root. The product of roots is c/a = (p-q)/(q-r). Since one root is 1, the other root is (p-q)/(q-r).
By substituting x equal to 1 into the expression (q minus r) times 1 squared plus (r minus p) times 1 plus (p minus q), the result is q minus r plus r minus p plus p minus q, which simplifies to 0. Since x equal to 1 is a verified root, the sum of the roots is negative (r minus p) divided by (q minus r), meaning the second root is (p minus q) divided by (q minus r). Therefore, (p minus q) divided by (q minus r) is a root of the equation.