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If $p$ and $q$ are zeroes of the polynomial $p(y) = 21y^2 - y - 2$, then find the value of $(1-p) . (1 - q)$.
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$p+q=\frac{1}{21}$
$p.q=\frac{-2}{21}$
$(1-p)(1 - q) = 1 - (p + q) + pq$
$= 1-\frac{1}{21} - \frac{2}{21}$
$= \frac{18}{21}$ or $\frac{6}{7}$
$p.q=\frac{-2}{21}$
$(1-p)(1 - q) = 1 - (p + q) + pq$
$= 1-\frac{1}{21} - \frac{2}{21}$
$= \frac{18}{21}$ or $\frac{6}{7}$