The roots of the given equation (p-q) x2+(q-r) x+(r-p)=0 are :

Question

The roots of the given equation (p-q) x2+(q-r) x+(r-p)=0 are : (A) ( p - q / r - p ), 1 (B) (q-r/p-q), 1 (C) (r-p/p-q), 1 (D) None of these

Tardigrade Admin · Accepted Answer

Given equation is (p-q) x2+(q-r) x+(r-p)=0 By using formula for finding the roots viz: (-b ± √b2-4 ac/2a), 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

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

$\frac{p - q}{r - p}, 1$

$\frac{q - r}{p - q}, 1$

$\frac{r - p}{p - q}, 1$

None of these

Q. The roots of the given equation (p−q)x2+(q−r)x+(r−p)=0 are :

r−pp−q​,1

p−qq−r​,1

p−qr−p​,1

None of these

Given equation is (p−q)x2+(q−r)x+(r−p)=0 By using formula for finding the roots viz : 2a−b±b2−4ac​​, we get x=2(p−q)(r−q)±(q−r)2−4(r−p)(p−q)​​ ⇒x=2(p−q)(r−q)±(q+r−2p)​=p−qr−p​,1

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

$\frac{p - q}{r - p}, 1$

$\frac{q - r}{p - q}, 1$

$\frac{r - p}{p - q}, 1$