For α, β ∈ ℝ , suppose the system of linear equations x - y + z = 5...

Given,

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

has infinitely many solutions,

So, finding $△ = 0$ we get,

\left|$\begin{cases}1 & - 1 & 1 \\ 2 & 2 & \alpha \\ 3 & - 1 & 4\end{cases}$\right| = 0$</p><p>$⇒ 8 + \alpha - 2 ( - 4 + 1 ) + 3 ( - \alpha - 2 ) = 0$</p><p>$⇒ 8 + \alpha + 6 - 3\alpha - 6 = 0$</p><p>$⇒ \alpha = 4$</p><p>Now finding${△}_{z} = 0we get,</p><p>\left|\begin{cases}1 & - 1 & 5 \\ 2 & 2 & 8 \\ 3 & - 1 & \beta\end{cases}\right| = 0

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

$⇒ \beta = 14$

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

$⇒ {x}^{2} - 18x + 56 = 0$