Question

4x²-17× + 4 =0 formula de bhaskara

Answer 1

$\Large\mathsf\displaystyle{} {4x}^{2} - 17x + 4 = 0 \\\\ \Large\mathsf\displaystyle{}\begin{cases}a = 4 \\ b = - 17 \\ c = 4\end{cases} \\\\\Large\mathsf\displaystyle{}x = \dfrac{ - \left( - 17\right) \pm \sqrt{\left( - 17\right)^{2} - 4\cdot4\cdot4} }{2\cdot4} \\\\ \\ \Large\mathsf\displaystyle{}x = \dfrac{17 \pm \sqrt{289 - 64} }{8} \\\\ \\ \Large\mathsf\displaystyle{}x = \dfrac{17 \pm \sqrt{225} }{8} \\\\\\\Large\mathsf\displaystyle{}x = \dfrac{17 \pm15}{8} \\\\ \\ \Large\mathsf\displaystyle{}x = \dfrac{17 + 15}{8} \\ \\ \Large\mathsf\displaystyle{}x = \dfrac{17 - 15}{8} \\ \\ \\ \\ \Large\mathsf\displaystyle{}x = 4 \\ \\ \Large\mathsf\displaystyle{}x = \dfrac{1}{4}$

Solução:

$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{ x_{1} = \dfrac{1}{4} }}}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{ x_{2} = 4 \\ }}}\end{gathered} }$

$\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{{ \red{Bons}} \blue{\:Estudos}}}}\end{gathered} }$