Answers

2014-10-06T21:27:39+05:30
If -3 is a zero it should satisfy the equation
substitute -3 in the equation and equate it to 0
(k-1)x²+kx+1=0
⇒(k-1)(-3)²+k(-3)+1=0
⇒9k-9-3k+1=0
⇒6k-8=0
⇒k=4/3
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2014-10-06T21:32:54+05:30

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( k - 1) x^2 + k x + 1 = 0 \\ \\roots=\frac{-k +- \sqrt{k^2 - 4(k-1)}}{2(k-1)}=\frac{-k+- \sqrt{(k-2)^2}}{2(k - 1)}\\ \\-\frac{-k+-(k-2)}{2(k-1)}\\ \\= -\frac{(2-2k)}{2(k-1)}\ \ or,\ \ \frac{-2}{2(k-1)}\\ \\=-1\ or\ -1/(k-1)\\ \\So,\ \ - 3 = -\frac{1}{k-1}\\ \\3k -3 = 1\\ \\k = 4/3\\

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Another way, perhaps more simple too:

When -3 is a root of a polynomial P(x) = (k-1)x^2 + k x + 1,\ \ then\ \ P(-3)=0\\ \\(k-1)\ (-3)^2 + k \ (-3) + 1 = 0 \\ \\ 9k-9-3k+1 = 0\\ \\6k-8=0\\k=4/3\\
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