Question

Resolva as equções abaixo admitindo raízes complexas
a) x²-6x+10 = 0 b) x²-2x+26= 0

Answer 1

Lembrando do produto notável (a-b)² = a²-2ab+b²

Completando os quadrados

$\begin{array}{l}\mathsf{x^2-6x+10=0}\\\\\mathsf{x^2-6x=-10}\\\\\mathsf{x^2-6x+9=-1}\\\\\mathsf{(x-3)^2=-1}\\\\\mathsf{\sqrt{(x-3)^2}=\sqrt{-1}}\\\\\mathsf{x-3=\pm~\sqrt{i^2}}\\\\\mathsf{x-3=\pm~i}\\\\\\\mathsf{S=}\begin{Bmatrix}\mathsf{x'=3-i}\\\\\mathsf{x''=3+i}\end.\end{array}$

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$\begin{array}{l}\mathsf{x^2-2x+26=0}\\\\\mathsf{x^2-2x=-26}\\\\\mathsf{x^2-2x+1=-25}\\\\\mathsf{(x-1)^2=-25}\\\\\mathsf{\sqrt{(x-1)^2}=\sqrt{-25}}\\\\\mathsf{x-1=\pm~\sqrt{25\cdot i^2}}\\\\\mathsf{x-1=\pm~5i}\\\\\\\mathsf{S=}\begin{Bmatrix}\mathsf{x'=1-5i}\\\\\mathsf{x''=1+5i}\end.\end{array}$

Answer 2

$x^{2} -6x+10=0$
Vamos usar a fórmula de bhaskara para resolver a equação do 2° grau.
$\Delta=\ b^{2}-4.a.c$
Vamos identificar A, B e C
$a = 1$
$b = -6$
$c = 10$

Vamos substituir os valores na fórmula
$\Delta\ = 6^{2} -4.1.10$
$\Delta\ = 36 - 40$
$\Delta\ = -4$

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

$\dfrac{-(-6)\pm\sqrt{-4}}{2.1}$

$\dfrac{6\pm\sqrt{-4}}{2}$

$x=\dfrac{6\pm\44i}{2}$

$\large\begin{array}{l}x=\dfrac{6\pm\ 4i}{2} = \frac{6}{2}\pm\ \frac{2i}{2}\end{array}$

$\large\begin{array}{l}S=\left \{ {{x' = 3+i}\atop {x'' = 3-i}} \right\end{array}$



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$x^{2} -2x+26=0$
$a = 1$
$b = -2$
$c = 26$

$\Delta=\ b^{2}-4.a.c$
$\Delta=-2^{2} -4.1.26$
$\Delta=\44-4.1.26$
$\Delta=\ 4-104$
$\Delta=\ -100$

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

$\dfrac{-(-2)\pm\sqrt{\ -100}}{2.1}$

$\dfrac{2\pm\sqrt{\ -100}}{2}$

$\large\begin{array}{l}S = \left \{ {{x'=1+5i} \atop {x''=1-5i}} \right\end{array}$