Question

the roots of the equation(p-q)x2 +(q-r)x+(r-p)=0 will be

Answer 1

(p-q)x^2+(q-p)x+(p-r)x+(r-p)=0
(p-q)x(x-1)+(p-r)(x-1)=0
(x-1)((p-q)x+(p-r))=0

so roots are
x=1 & x=(r-p)/(p-q)

Answer 2

Answer:

Option C is the right answer.

Step-by-step explanation:

On solving the given equations the roots can be obtained. Usually the roots are determined by their x-intercepts.

Given equation $\rightarrow(p-q) x^{2}+(q-r) x+(r-p)=0$

This equation can be written as:

$(p-q) x^{2}+(q-r) x+(r-p)=0$

On solving,

$\begin{array}{l}{\Rightarrow(p-q) x^{2}+(q x-r x)+(r-p)=0} \\ {=>(p-q) x^{2}+(q x-r x+p x-p x)+(r-p)=0} \\ {=>(p-q) x^{2}+(q x-p x+p x-r x)+(r-p)=0} \\ {=>(p-q) x^{2}+(q-p) x+(p-r) x+(r-p)=0} \\ {=>(p-q) x^{2}-(p-q) x-(r-p) x+(r-p)=0} \\ {=>(p-q) x(x-1)-(r-p)(x-1)=0} \\ {=>\{(p-q) x-(r-p)\}(x-1)=0}\end{array}$

Now comparing the equations,

Either,        Or,

{(p-q)x-(r-p)}=0             (x-1)=0

=>(p-q)x=(r-p)                    =>x=1

=>x=(r-p)/(p-q)          

The values of x are the roots.

Hence, $\frac{r-p}{p-q}$ and 1  are the roots. Option C.