Question

Obtenha os zeros da função

y= f(x) = -3x^2 + 5x + 2


y= f(x) = 2x^2 - 5x - 3


y= f(x) = x2 + 1

Answer 1

Oi!

a)
$f(x) = - 3x^2 + 5x + 2$

$\Delta = b^2 - 4ac$
$\Delta = 5^2 - 4 * (-3) * 2$
$\Delta = 25 + 24$
$\Delta =49 > 0$
(Há duas raízes reais distintas)

$x = \frac{ - b \pm \sqrt{\Delta}}{2a}$
$x = \frac{ - (5) \pm \sqrt{49}}{2(-3)}$
$x = \frac{ - 5 \pm 7}{-6}$

$x' = \frac{ - 5 + 7}{-6} = \frac{2}{-6} => x' = - \frac{1}{3}$

$x" = \frac{ - 5 - 7}{-6} x" = \frac{ - 12}{-6} => x" = 2}$

S = {- 1/3, 2}

b)$f(x) = 2x^2 - 5x + 2$

$\Delta = b^2 - 4ac$
$\Delta = (-5)^2 - 4 * 2 * (-3)$
$\Delta = 25 + 24$
$\Delta = 49$
(Há duas raízes reais distintas)

$x = \frac{ - b \pm \sqrt{\Delta}}{2a}$
$x = \frac{ - (-5) \pm \sqrt{49}}{2 * 2}$
$x = \frac{5 \pm 7}{4}$

$x' = \frac{5 + 7}{4} = \frac{12}{4} => x' = 3$

$x" = \frac{5 - 7}{4} = \frac{-2}{4} => x" = - \frac{1}{2}$

S = {-1/2, 3}

c)$f(x) = x^2 + 1$

$\Delta = b^2 - 4ac$
$\Delta = 0^2 - 4 * 1 * 1$
$\Delta = - 4 < 0$
(Não há raizes reais)

S = {}

Bons estudos!