Question

if one of the zeros of the quadratic polynomial

$(k - 1) {x}^{2} + kx + 1 \: is \: - 3 \: then \: the \: value \: of \: k \:$
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Answer 1

Hey Buddy ! ☺

$P(x)=(k - 1) {x}^{2} + kx + 1$

One zero = 3

Therefore;

$p(3) = (k - 1) ({-3})^{2} + 3k+ 1 \\ \\ = (k - 1)9 + 3k + 1 \\ \\ = 9k - 9 - 3k+ 1 \\ \\ = 6k - 8 \\ \\ \\ put \: p(x) = 0 \\ \\ = > 6k - 8 = 0 \\ \\ = > 6k = 8【divide \:eq\:by\:2】 \\ \\ = > 3k = 4\\ \\ = > \fbox{ k = \frac{4}{3} }$

Answer 2

     

$p(-3)=(k-1)(-3)^2+k(-3)+1=0 \\ \\ p(-3)=9(k-1)-3k+1=0 \\ \\ p(-3)=9k-9-3k+1=0 \\ \\ p(-3)=6k-8=0 \\ \\ \\ 6k-8=0 \\ \\ 6k=8 \\ \\ k=\frac{8}{6} \\ \\ \\ \therefore\ k=\bold{\frac{4}{3}}$

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