Question

2x+1=3x
$5$
ok

Answer 1

Oi!

$2x^2 + 1 = 3x$
$2x^2 - 3x + 1 = 0$

Calculando o Delta:

$\Delta = b^2 - 4ac$
$\Delta = (-3)^2 - 4 * 2 * 1$
$\Delta = 9 - 8$
$\Delta = 1 > 0$
(Há duas raízes reais e distintas)


Calculando as raízes:
$x = \frac{ - b \pm \sqrt{\Delta}}{2a}$
$x = \frac{ - (-3) \pm \sqrt{1}}{2 * 2}$
$x = \frac{ 3 \pm 1}{4}$

$x' = \frac{3 + 1}{4} = \frac{4}{4} => x'= 1$
$x'' = \frac{3 - 2}{4} = \frac{2}{4} => x" = \frac{1}{2}$

Soma e produto:
$S = x' + x'' = 1 + \frac{1}{2} = \frac{3}{2}$
$P = x' * x" = 1 * \frac{1}{2} = \frac{1}{2}$