Question

If 1 zero of (k-1) x square+(k-2) X+1 is negative of other then find k

Answer 1

${\huge{\pink{\underline{\underline{\purple{\mathbb{\bold{\mathcal{AnSwEr..}}}}}}}}}$

${\large{\blue{\bold{\underline{Given:-}}}}}$

• Polynomial,

$\sf\implies (k-1)x^{2}+(k-1)x+1$

• One zeroe is negative of another.

▪︎ Let the one zero be $\sf \alpha$

▪︎ So Another zero = $\sf -\alpha$

${\large{\blue{\bold{\underline{Concept\:Used:-}}}}}$

$\sf Sum\:Of\:Zeroes = \frac{-b}{a}$

$\sf\implies \alpha + ( - \alpha ) = \frac{ - (k - 2)}{(k - 1)} \\ \\ \sf\implies \alpha - \alpha = \frac{2 - k}{k - 1} \\ \\ \sf\implies0 = \frac{2 - k}{k - 1} \\ \\ \sf\implies0(k - 1) = 1(2 - k) \\ \\ \sf\implies2 - k = 0 \\ \\ \sf\implies \cancel- \: k = \cancel - \: 2 \\ \\ \sf\pink{\fbox{\implies \: k = 2}}$

☆ Therefore the required value of k is 2.