Question

Encontre tds as soluções reais

${x}^{2} - 9x = 36$
Resposta: {3,12}

${3x}^{2} = 2x + 8$
Resposta:
$- \frac{4}{3}$
-4/3,2


${4x}^{2} + 3x - 5 = 0$
Resposta:

{
$\frac{ - 3 + \sqrt{89} }{8}$
,
$\frac{ - 3 - \sqrt{89} }{8}$
}

{ela esta assim -3+2raiz de 89/8, -3-raiz de 89/8}



me ajudem, por favor.
​

Answer 1

A)

x²—9x=36

x²—9x—36=0

∆=(-9)²—4•1•(-36)

∆=81+144

∆=225

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2.a}$

$x_{1,2}=\frac{9\pm\sqrt{225}}{2.1}=\frac{9\pm\:15}{2}$

$x_{1}=\frac{9+15}{2}=\frac{24}{2}$

$x_{1}=12$

$x_{2}=\frac{9-15}{2}=\frac{-6}{2}$

$x_{2}=-3$

Sol: { -3 , 12 }

C)

3x²=2x+8

3x²—2x—8=0

∆=(-2)²—4•3•(-8)

∆=4+96

∆ = 100

$x_{1,2}=\frac{2\pm\sqrt{100}}{2.3}=\frac{2\pm\:10}{6}$

$x_{1}=\frac{2+10}{6}=\frac{12}{6}$

$x_{1}=2$

$x_{2}=\frac{2-10}{6}=\frac{-8}{6}$

$x_{2}=\frac{-4}{3}$

Sol: { -4/3 , 2 }

C)

4x²+3x—5=0

∆ = 3²—4•4•(—5)

∆ = 9+80

∆ = 89

$x_{1,2}=\frac{-3\pm\sqrt{89}}{2.4}$

$x_{1,2}=\frac{-3\pm\sqrt{89}}{8}$

$Sol:\{\frac{-3\pm\sqrt{89}}{8}\}$

Espero ter ajudado bastante;)

Data: 23/06/2019