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Question

Using quadratic formula solve the following quadratic equation: [3 MARKS]

p2x2+(p2q2)xq2=0,p0

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Solution

Concept: 1 Mark
Application: 2 Marks


We have, p2x2+(p2q2)xq2=0,p0

Comparing this equation with ax2+bx+c=0, we have

a=p2,b=p2q2 and c=q2

D=b24ac=(p2q2)24×p2×q2

D=(p2q2)2+4p2q2

D=(p2+q2)2

D>0

So, the given equation has real roots given by

α=b+D2a=(p2q2)+(p2+q2)2p2=q2p2

and, β=bD2a=(p2q2)(p2+q2)2p2=1


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