Using quadratic formula solve the following quadratic equation: [3 MARKS]
p2x2+(p2−q2)x−q2=0,p≠0
Concept: 1 Mark
Application: 2 Marks
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
∴D=b2−4ac=(p2−q2)2−4×p2×−q2
⇒D=(p2−q2)2+4p2q2
⇒D=(p2+q2)2
⇒D>0
So, the given equation has real roots given by
α=−b+√D2a=−(p2−q2)+(p2+q2)2p2=q2p2
and, β=−b−√D2a=−(p2−q2)−(p2+q2)2p2=−1