Question

2x²+3x+1 find the zeroes of the following quadratic polynomial in simple form​

Answer 1

$\\ \large\sf{\orange{«\: คꈤ \mathfrak Sฬєя \: » }} \\ \\ \begin{gathered}⇛2 {x}^{2} - 3x + 1 = 0 \\ \\⇛ 2 {x}^{2} - 2x - x + 1 = 0 \\ \\ ⇛2x(x - 1) - 1(x - 1) = 0 \\ \\ ⇛(x - 1)(2x - 1) = 0 \\ \\ ⇛\alpha = 1 \\ \\⇛ \beta = \frac{1}{2} \\ \\ ⇛3 \alpha = 3 \times 1 = 3 \\ \\ ⇛3 \beta = 3 \times \frac{1}{2} = \frac{3}{2} \\ \\ \sf{so \: polynomial} \\ \\ ⇛{x}^{2} - (3 \alpha + 3 \beta )x + (3 \alpha )(3 \beta ) = 0 \\ \\ ⇛{x}^{2} - (3 + \frac{3}{2} )x + (3)( \frac{3}{2} ) = 0 \\ \\ ⇛{x}^{2} - \frac{9}{2} x + \frac{9}{2} = 0 \\ \\⇛ 2 {x}^{2} - 9x + 9 = 0\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \huge \bold{@WildCat7083 }$