Question

The root of question(q-r)x2+ (r-p)x+(p-q)=0 are ?

Answer 1

Question:

Find the roots of the given equation;

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

Answer:

x = 1 and (p-q)/(q-r)

Solution:

We have;

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

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

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

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

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

=> (q-r)x(x-1) - (p-q)(x-1) = 0

=> (x-1){(q-r)x - (p-q)} = 0

Case(1):

If (x-1) = 0

=> x = 1

Case(2):

If {(q-r)x - (p-q)} = 0

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

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

or x = (q-p)/(r-q)

Hence,

The roots of the given equation are :

x = 1 and (p-q)/(q-r)

Answer 2

Question :-

The roots of this quadratic equation are -

$\small \: \green{(q - r) {x}^{2} + (r - p)x + (p - q) = 0}$

Answer:-

$\implies \: \green{\: 1 \: } \: \: and \: \red{\frac{(p - q)}{(q - r)} }$

Step by step explanation :-

Given quadratic equation is

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

We can write ,

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

$\small \: (q - r)x(x - 1) - (p - q)(x - 1) = 0 \\ \\ \small \: (x - 1)[(q - r)x - (p - q)] = 0$

$\star \: case \: (1) \\ \\ \implies \: \small \: (x - 1) = 0 \\ \\ \implies \: \ \red{\boxed{\small x = 1 }}\\ \\ \star \: case \: (2) \: \\ \\ \implies \: \red{\small \: (q - r)x - (p - q) = 0 }\\ \\ \implies \: \small \: (q - r)x = (p - q) \\ \\ \implies \: \red{ \boxed{ \small \: x = \frac{(p - q)}{(q - r)} }}$