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
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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