Date: खाली म्यान हित
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Given: p(x)=ax
2
−3(a−1)x−1
One of the zero =1
As one of the zeroes is 1, substituting it in the polynomial
⇒p(1)=0=a−3(a−1)−1
⇒0=−2a+3−1
⇒a=1
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