Question

qual o valor de x para que √x2-x+4 seja 4

Answer 1

$\sqrt{ {x}^{2} - x +4} = 4 \\ { \sqrt{x {}^{\cancel{2}} - x + 4} } ^{\cancel{2}} = {4}^{2} \\ {x}^{2} - x + 4 = 16$

Faça a equação do segundo grau

${x}^{2} - x + 4 - 16 = 0 \\ {x}^{2} - x - 12 = 0\begin{gathered}\begin{cases} {a = 1}\\{b = - 1} \\ c = - 12\end{cases}\end{gathered}$

$Δ = {b}^{2} - 4ac \\ Δ = ( - 1 {)}^{2} - 4 \times 1 \times ( - 12) \\ Δ = 1 +48 \\ Δ = 49$

$x = \frac{ - b± \sqrt{Δ} }{2a} \\ \\ x = \frac{ - ( - 1)± \sqrt{49} }{2 \times 1} \\ \\ x = \frac{1±7}{2} \begin{gathered}\begin{cases} { \frac{1 + 7}{2} = \frac{8}{2} = 4}\\{ \frac{1 - 7}{2} = \frac{ - 6}{2} = - 3 } \end{cases}\end{gathered}$

$x_{1} = - 3 \\ x_{2} = 4$

Prova final:

$\sqrt{( - 3 {)}^{2} - ( - 3) + 4 } = 4 \\ \sqrt{9 + 3 + 4} = 4 \\ \sqrt{16 } = 4 \\ 4 = 4$

$\sqrt{ {4}^{2} - 4 + 4 } = 4 \\ \sqrt{ {4}^{2} \cancel{ - 4 + 4}} = 4 \\ \sqrt{ {4}^{2} } = 4 \\ \sqrt[\cancel{2}]{ {4}^{\cancel{2}} } = 4 \: \: \: \: \: \: ou \: \: \: \: \sqrt{ {4}^{2} }= \sqrt{16} = 4 \\ 4 = 4\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 4 = 4$

$\mathcal{Bons \: estudos } \\ \displaystyle\int_ \empty ^ \mathbb{C}     \frac{ - b \: ± \:  \sqrt{ {b}^{2} - 4 \times a \times c } }{2 \times a} d{ t } \boxed{ \boxed{ \mathbb{\displaystyle\Re}\sf{ \gamma  \alpha }\tt{ \pi}\bf{ \nabla}}}$