Question

Gostaria de saber a resposta da 1!

Answer 1

a) x^2 - 10x + 9 = 0

Fórmula de Bhaskara
b^2 - 4 . a . c
(-10)^2 - 4 . 1 . 9
100 - 36
Delta = 64

Fórmula X1 e X2
X1 = -b + √64 ÷ 2 . a
X1 = 10 + 8 ÷ 2 . 1
X1 = 18 ÷ 2
X1 = 9

X2 = 10 - 8 ÷ 2
X2 = 2 ÷ 2
X2 = 1
____________________

b) x^2 - 6x + 5 = 0

(-6)^2 - 4 . 1 . 5
36 - 20
Delta = 16

X1 = 6 + 4 ÷ 2 . 1
X1 = 10 ÷ 2
X1 = 5

X2 = 6 - 4 ÷ 2 . 1
X2 = 2 ÷ 2
X2 = 1
____________________

c) x^2 - 10x + 25

(-10)^2 - 4 . 1 . 25
100 - 100
Delta = 0

X1 = 10 + 0 ÷ 2 . 1
X1 = 10 ÷ 2
X1 = 5

X2 = 10 - 0 ÷ 2 . 1
X2 = 10 = 2
X2 = 5

Answer 2

$1)\\ \\ a)\\ \\ x^{2} -10x+9=0$

Δ = b² - 4.a.c

Δ = -10² - 4.1.9

Δ = 100 - 36

Δ = 64

$x=\frac{-b+-\sqrt{delta} }{2a} \\ \\ x=\frac{10+-\sqrt{64} }{2} \\ \\ x=\frac{10+-8}{2} \\ \\ x'=\frac{10+8}{2} =\frac{18}{2} =9\\ \\ x''=\frac{10-8}{2}=\frac{2}{2} =1\\ \\ S=(1;~~9)$

$b)\\ \\ x^{2} -6x+5=0$

Δ = b² - 4.a.c

Δ = -6² - 4.1.5

Δ = 36 - 20

Δ = 16

$x=\frac{-b+-\sqrt{delta} }{2a} \\ \\ x=\frac{6+-\sqrt{16} }{2} \\ \\ x=\frac{6+-4}{2} \\ \\x'=\frac{6+4}{2} =\frac{10}{2} =5\\ \\ x''=\frac{6-4}{2} =\frac{2}{2} =1\\ \\ S=(5;~~1)$

$c)\\ \\ x^{2} -10x+25=0$

Δ = b² - 4.a.c

Δ = -10² - 4.1.25

Δ = 100 - 100

Δ = 0

Quando Δ (delta) = 0 (zero) aplique a fórmula de Bhaskara, reduzida:

$x=\frac{-b}{2a} \\ \\ x=\frac{10}{2} \\ \\ x=5$