Question

fórmula de báscara...
preciso do desenvolvimento!!!
a)x² -8x +15 = 0
b)x² +10x +25 = 0
c)3x² +4x +1 = 0
d)-x² +12x -20 = 0
e)x² -8x +12 = 0

Answer 1

a)
$x^{2} -8x +15 = 0$

Encontrar o Delta (Δ)

Δ=b2−4ac
Δ=(−8)2−4*(1)*(15)
Δ=64−60
Δ=4

$x = \frac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \frac{-(-8) \pm \sqrt{4}}{2*1} \\ \\x = \frac{8 \pm {2}}{2} \\ \\ x' = \frac{8 + {2}}{2} \\ \\ x' = \frac{10}{2} => x' = 5 \\ \\ x'' = \frac{8 - {2}}{2} \\ \\ x'' = \frac{6}{2} => x'' = 3$

S= {5, -3}

b) 
$x^{2} +10x +25 = 0$

Δ=b2−4ac
Δ=(10)2−4*(1)*(25)
Δ=100−100
Δ=0

$x = \frac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \frac{-10 \pm \sqrt{0}}{2*1} \\ \\x = \frac{-10 \pm {0}}{2} \\ \\ x' = \frac{-10 + {0}}{2} \\ \\ x' = \frac{-10}{2} => x' = -5 \\ \\ x'' = \frac{-10 - {0}}{2} \\ \\ x'' = \frac{-10}{2} => x'' = -5$

S = {-5}

c)
$3 x^{2} +4x +1 = 0$

Δ=b2−4ac
Δ=(4)2−4*(3)*(1)
Δ=16−12
Δ=4
$x = \frac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \frac{-4 \pm \sqrt{4}}{2*3} \\ \\x = \frac{-4 \pm {2}}{6} \\ \\ x' = \frac{-4 + {2}}{6} \\ \\ x' = \frac{-2}{6} => x' = - \frac{2}{6} => x' = -\frac{1}{3} \\ \\ x'' = \frac{-4 - {2}}{6} \\ \\ x'' = \frac{-6}{6} => x'' = -1$

S = (-$\frac{1}{3} , -1$)

d)
$- x^{2} +12x -20 = 0$    vamos multiplicar por (-1)

$x^{2} -12x +20 = 0$

Δ=b2−4ac
Δ=(−12)2−4*(1)*(20)
Δ=144−80
Δ=64

$x = \frac{-(-12) \pm \sqrt{64}}{2*1} \\ \\x = \frac{12 \pm {8}}{2} \\ \\ x' = \frac{12 + {8}}{2} \\ \\ x' = \frac{20}{2} => x' = 10 \\ \\ x'' = \frac{12 - {8}}{2} \\ \\ x'' = \frac{4}{2} => x'' = 2$

S= {10,2}

e)
$x^{2} -8x +12 = 0$

Δ=b2−4ac
Δ=(−8)2−4*(1)*(12)
Δ=64−48
Δ=16

$x = \frac{-(-8) \pm \sqrt{16}}{2*1} \\ \\x = \frac{8 \pm {4}}{2} \\ \\ x' = \frac{8 + {4}}{2} \\ \\ x' = \frac{12}{2} => x' = 6 \\ \\ x'' = \frac{8 - {4}}{2} \\ \\ x'' = \frac{4}{2} => x'' = 2$

S ={6, 2}