For α , β ∈ ℝ , suppose the system of linear equations x - y + z = 5 2 x + 2 y + &#945…

Given,

$\alpha ,\beta \in \mathrm{ℝ}$, suppose the system of linear equations
$x-y+z=5$ 
$2x+2y+\alpha z=8$
$3x-y+4z=\beta$

has infinitely many solutions,

So, finding $△=0$ we get,

$\left|\begin{array}{ccc}1& -1& 1\\ 2& 2& \alpha \\ 3& -1& 4\end{array}\right|=0$

$\Rightarrow 8+\alpha -2(-4+1)+3(-\alpha -2)=0$

$\Rightarrow 8+\alpha +6-3\alpha -6=0$

$\Rightarrow \alpha =4$

Now finding ${△}_z=0$ we get,

$\left|\begin{array}{ccc}1& -1& 5\\ 2& 2& 8\\ 3& -1& \beta \end{array}\right|=0$

$\Rightarrow 3\left(-18\right)+1\left(-2\right)+\beta \left(4\right)=0$

$\Rightarrow \beta =14$

Hence, quadratic equation having roots $4 \&14$ will be, $x^2-\left(4+14\right)x+4\times 14=0$

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