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प्रश्न
Using quadratic formula find the value of x.
p2x2 + (p2 – q2)x – q2 = 0
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उत्तर
We have p2x2 + (p2 – q2)x – q2 = 0
Comparing this equation to ax2 + bx + c = 0, we have
a = p2, b = p2 – q2 and c = – q2
∴ D = b2 – 4ac
= (p2 – q2)2 – 4 × p2 + (– q)2
= (p2 – q2)2 + 4p2q2
= (p2 + q2)2 > 0
So, the given equation has real roots given by
`alpha = (-b + sqrt(D))/(2a)`
= `(-(p^2 - q^2) + (p^2 + q^2))/(2p^2)`
= `(2q^2)/(2p^2)`
= `q^2/p^2`
And `beta = (-b - sqrt(D))/(2a)`
= `(-(p^2 - q^2) - (p^2 + q^2))/(2p^2)`
= `(-2q^2)/(2p^2)`
= –1
Hence roots are `q^2/p^2` and –1
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