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Question

Using quadratic formula, find the value of x.
p2x2+(p2q2)xq2=0,p0

A
q2p2,1
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B
q2p2,1
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C
0,1
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D
q2p2,1
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Solution

The correct option is A q2p2,1
We have, p2x2+(p2q2)xq2=0, p0

Comparing this equation with ax2+bx+c=0, we have,
a=p2,b=p2q2 and c=q2.
Now, discriminant,
b24ac=(p2q2)24×p2×q2

=(p2q2)2+4p2q2

=(p2+q2)2

Now finding roots,

x=b±b24ac2a

x=(p2q2)±(p2+q2)22p2

x=(p2q2)+(p2+q2)2p2 or x=(p2q2)(p2+q2)2p2

x=q2p2 or x=1


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