Question : $x$ is a negative number such that $k+k^{-1}=-2$, then what is the value of $\frac{k^2+4 k-2}{k^2+k-5}$?
Option 1: 7
Option 2: 1
Option 3: –7
Option 4: –1
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Correct Answer: 1
Solution : Given: $k+k^{-1}=-2$ ⇒ $k+\frac{1}{k}=-2$ ⇒ $k^2+1=-2k$ ⇒ $k^2+2k+1=0$ ⇒ $(k + 1)^2=0$ $\therefore k=-1$ Putting the value of $k$ in the given equation, $\frac{k^2+4 k-2}{k^2+k-5}$ $=\frac{(-1)^2+4 (-1)-2}{(-1)^2+(-1)-5}$ $=\frac{1-4-2}{1-1-5}$ $=\frac{-5}{-5}$ $=1$ Hence, the correct answer is 1.
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