Using quadratic formula, find the value of x.
p2x2+(p2−q2)x−q2=0,p≠0
The correct option is A q2p2,−1
We have, p2x2+(p2−q2)x−q2=0, p≠0
Comparing this equation with ax2+bx+c=0, we have,
a=p2,b=p2−q2 and c=−q2.
Now, discriminant,
b2−4ac=(p2−q2)2−4×p2×−q2
=(p2−q2)2+4p2q2
=(p2+q2)2
Now finding roots,
x=−b±√b2−4ac2a
⇒x=−(p2−q2)±√(p2+q2)22p2
⇒x=−(p2−q2)+(p2+q2)2p2 or x=−(p2−q2)−(p2+q2)2p2
⇒x=q2p2 or x=−1