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For \alpha, \beta \in \mathbb{R}, suppose the system of linear equations
\mathrm{x}-\mathrm{y}+\mathrm{z}=5
2 x+2 y+\alpha z=8
3 x-y+4 z=\beta
has infinitely many solutions. Then \alpha$ and $\beta are the roots of

Option: 1

x^{2}+14 x+24=0


Option: 2

x^{2}+18 x+56=0


Option: 3

x^{2}-18 x+56=0


Option: 4

x^{2}-10 x+16=0


Answers (1)

\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{array}\right|=0
1(8+\alpha)+1(8-3 \alpha)+1(-2-6)=0
\Rightarrow 8+\alpha+8-3 \alpha-8=0
-2 \alpha=-8
\alpha=4
D_{1}=0

\Rightarrow\left|\begin{array}{lll} 5 & -1 & 1 \\ 8 & 2 & 4 \\ \beta & -1 & 4 \end{array}\right|=0
5(8+4)+1(32-4 \beta)+1(-8-2 \beta)=0
60+32-4 \beta-8-2 \beta=0
\Rightarrow-6 \beta=-84
\beta=14

Equation having roots as \alpha \& \beta
x^{2}-18 x+56=0

Posted by

Sumit Saini

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