Question

Determine as raízes das equações a seguir:

Answer 1

As soluções são 3 ou 2.

x² - 5x + 6  = 0

a = 1

b = 5

c = 6

$x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$

$x=\dfrac{-(-5)\pm\sqrt{\Delta}}{2}$

∆ = b² - 4ac

∆ = -5² - 4 * 1 * 6

∆ = 25 - 4 * 1  * 6

∆ = 25 - 4 * 6

∆ = 25 - 24

∆ = 1

$x=\dfrac{5\pm\sqrt{1}}{2}$

$\sqrt{1} = 1$

$x=\dfrac{5\pm{1}}{2}$

$x_{1} = \dfrac{5 + 1}{2}$

$x_{1} = \dfrac{6}{2}$

$x_{1} = 3$

$x_{2} = \frac{5 - 1}{2}$

$x_{2} = \dfrac{4}{2}$

$x_2 = 2$

Answer 2

Resposta:

$\sf x^{2} - 5x + 6 = 0$

$\sf ax^{2} +bx + c = 0$

a = 1

b = - 5

c = 6

Determinar o Δ:

$\sf \Delta = b^2 -\:4ac$

$\sf \Delta = (-5)^2 -\:4 \cdot 1 \cdot 6$

$\sf \Delta = 25 -\: 24$

$\sf \Delta = 1$

Determinar as raízes da equação:

$\sf x = \dfrac{-\,b \pm \sqrt{ \Delta } }{2a} = \dfrac{5 \pm \sqrt{ 1 } }{2\cdot 1} = \dfrac{5\pm1}{2} \Rightarrow\begin{cases} \sf x_1 = &\sf \dfrac{5 + 1}{2} = \dfrac{6}{2} = 3 \\\\ \sf x_2 = &\sf \dfrac{5 - 1}{2} = \dfrac{4}{2} = 2\end{cases}$

$\sf \boldsymbol{ \sf \displaystyle S = \{ x \in \mathbb{R} \mid x = 2 \mbox{\sf \;e } x = 3 \} }$