Question

( x - 1/4)^2 = 9/16.
Fórmula de bháskara.
Algem poderia me ajudar

Answer 1

Olá, bom dia =D

$(x - \frac{1}{4})^2 = \frac{9}{16} \\\\ (x - \frac14)(x - \frac14) = \frac{9}{16} \\\\ x^2 -\frac14x - \frac14x + \frac{1}{16} = \frac{9}{16} \\\\ x^2 - \frac24x + \frac{1}{16} - \frac{9}{16} = 0 \\\\ x^2 - \frac24x - \frac{8}{16}= 0 \qquad --\ \textgreater \ Simplificando~as~fracoes\\\\ x^2 - \frac12x - \frac{1}{2}= 0$

Agora vamos para a parte de bhaskara:

$Retirando~informacoes:\\\\ 1x^2-\frac12x-\frac12 = 0\\ ax^2+bx+c=0\\\\ a =1\\ b=-\frac12 \\ c=-\frac12 \\\\ Resolvendo: \\\\ \Delta=b^2-4.a.c\\\\ \Delta=(-\frac12)^2 - 4.1.(-\frac12)\\\\ \Delta=\frac14 + 2\\\\\ \Delta=\frac94\\\\\\\\ x^+=\frac{-b \pm \sqrt{\Delta}}{2.a}\\\\ x^+=\frac{-(-\frac12)+\sqrt{\frac{9}{4}} }{2.1}$

$x^+ = \frac{\frac12+\frac32}{2} \qquad \qquad\qquad x^-= \frac{\frac12 - \frac32}{2} \\\\ x^+ = \frac{\frac42}{2} \qquad \qquad\qquad~~~ x^-= \frac{-\frac22}{2} \\\\ x^+ = \frac{4}{2.2} \qquad \qquad\qquad~~ x^- = -\frac{1}{2} \\\\ x^+ = 1 \qquad \qquad\qquad~~~~ x^- = -0,5$

Nuss, que trabalheira, espero ter ajudado.
Bons estudos.