Question

como resolver a equacao oito x ao quadrado menos seis x mais um igual a zero??

Answer 1

Por Báskara:

x=(-b+-sqrt(delta=b²-4*a*c))/2*a

Colocando na fórmula:

x=(-(-6)+-sqrt((-6)²-4*8*1))/2*8

x=(6+-sqrt(4))/16

x=(6+-2)/16

x'=(6+2)/16=  1/2

x"=(6-2)/16= 1/4

Answer 2

Forma 1. (Fatoração por agrupamento)

$8x^{2}-6x+1=0\\ \\ 8x^{2}-4x-2x+1=0\\ \\ 4x\cdot (2x-1)-1\cdot (2x-1)=0\\ \\ (2x-1)(4x-1)=0\\ \\ \begin{array}{c} \begin{array}{rcl} 2x-1=0&~\text{ ou }~&4x-1=0\\ \\ 2x=1&~\text{ ou }~&4x=1\\ \\ \end{array}\\ \boxed{\begin{array}{rcl} x=\dfrac{1}{2}&~\text{ ou }~&x=\dfrac{1}{4} \end{array}} \end{array}$


Forma 2. (Fórmula resolutiva de Báscara)

$8x^{2}-6x+1=0~~~\Rightarrow~~\left\{ \begin{array}{l} a=8\\b=-6\\c=1 \end{array} \right.\\ \\ \\ \Delta=b^{2}-4ac\\ \\ \Delta=(-6)^{2}-4\cdot 8\cdot 1\\ \\ \Delta=36-32\\ \\ \Delta=4$

$x=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\ \\ \\ x=\dfrac{-(-6)\pm \sqrt{4}}{2\cdot 8}\\ \\ \\ x=\dfrac{6\pm 2}{2\cdot 8}\\ \\ \\ x=\dfrac{\diagup\!\!\!\! 2\cdot (3\pm 1)}{\diagup\!\!\!\! 2\cdot 8}\\ \\ \\ x=\dfrac{3\pm 1}{8}\\ \\ \\ \begin{array}{c} \begin{array}{rcl} x=\dfrac{3-1}{8}&~\text{ ou }~&x=\dfrac{3+1}{8}\\ \\ x=\dfrac{2}{8}&~\text{ ou }~&x=\dfrac{4}{8}\\ \\ x=\dfrac{\diagup\!\!\!\! 2}{\diagup\!\!\!\! 2\cdot 4}&~\text{ ou }~&x=\dfrac{\diagup\!\!\!\! 4}{\diagup\!\!\!\! 4\cdot 2}\\ \\ \end{array}\\ \boxed{\begin{array}{rcl} x=\dfrac{1}{4}&~\text{ ou }~&x=\dfrac{1}{2} \end{array}} \end{array}$