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If the system of simultaneous linear equations $x+y-z=6,4 x+y+z=2$ and $x+k y+z=-8$ has a unique solution $x=2$, $y=\beta, z=\gamma$, then the value of $k$ satisfies the following quadratic equation
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The correct answer is:
$x^2+x-2=0$
$x+y-z=6$
$\begin{aligned} & 4 x+y+z=2 \\ & x+k y+z=-8\end{aligned}$
$x=2, y=\beta, z=\gamma$
$y-3=4 \quad$ ...(i)
$y+z=-10$ ...(ii)
$2 y=-6$
$\Rightarrow y=-3$ Put in eq no (i)
$3=-7$
Now, put values of $x, y$ and $z$, we get
$2+k(-3)+(-7)=-8$
$\Rightarrow-3 k=-3$
$\Rightarrow k=1 \longrightarrow$ check all the options which satisfy this value of $k$.
we get option (d),
$\begin{aligned} & x^2+x-2=0 \\ & \Rightarrow(1)^2+(1)-2=0 \Rightarrow 0=0 \text { verified. }\end{aligned}$
$\begin{aligned} & 4 x+y+z=2 \\ & x+k y+z=-8\end{aligned}$
$x=2, y=\beta, z=\gamma$
$y-3=4 \quad$ ...(i)
$y+z=-10$ ...(ii)
$2 y=-6$
$\Rightarrow y=-3$ Put in eq no (i)
$3=-7$
Now, put values of $x, y$ and $z$, we get
$2+k(-3)+(-7)=-8$
$\Rightarrow-3 k=-3$
$\Rightarrow k=1 \longrightarrow$ check all the options which satisfy this value of $k$.
we get option (d),
$\begin{aligned} & x^2+x-2=0 \\ & \Rightarrow(1)^2+(1)-2=0 \Rightarrow 0=0 \text { verified. }\end{aligned}$
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