Question

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

• A. $x^2-18 x+56=0$
• B. $x^2+14 x+24=0$
• C. $x^2-10 x+16=0$
• D. $x^2+18 x+56=0$

Answer 1

Answer: (A)

$x^2-18 x+56=0$

Answer 2

$\begin{vmatrix} 1 & -1 & 1\\\ 2 & 2& \alpha\\\3&-1&4\end{vmatrix}$
= $0;8+α−2(−4+1)+3(−α−2)=0$
$\Rightarrow 8+α+6−3α−6=0$
$α=4$
So, the correct Option is (A): $x^2-18 x+56=0$