Question

Utilizando o processo algébrico de bhaskara,determine as raizes das equações do 2º grau no conjunto dos números reais :
a) x²-10x + 9 = 0
b) x² + x - 6 = 0
c) x² + 4x- 5 = 0
d) x²-10x + 24 = 0
e)2x²-9x+4 = 0
f) x²+8x+16 = 0

Answer 1

$x = \frac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

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

a=1, b=−10, c=9
Δ=b2−4ac
Δ=(−10)2−4*(1)*(9)
Δ=100−36
Δ=64

$\\ \\ x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-10) \pm \sqrt{64}}{2*1} \\ \\ x = \frac{10 \pm 8}{2} \\ \\ x' = \frac{10 + 8}{2} \\ \\ x' = \frac{18}{2} \\ \\ x' = 9 \\ \\ x'' = \frac{10 - 8}{2} \\ \\ x'' = \frac{2}{2} \\ \\ x'' = 1$

S = {9, 1}
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$b)\\ x^2 + x - 6 = 0$

a=1, b=1, c=−6
Δ=b2−4ac
Δ=(1)2−4*(1)*(−6)
Δ=1+24
Δ=25
$x = \frac{-b \pm \sqrt{\triangle}}{2*a}\\ \\x = \frac{-1 \pm \sqrt{25}}{2*1}\\ \\x = \frac{-1 \pm 5}{2} \\ \\ x' = \frac{-1 + 5}{2}\\ \\x' = \frac{4}{2}\\ \\x' = 2\\ \\ \\x'' = \frac{-1 - 5}{2}\\ \\x'' = \frac{-6}{2}\\ \\x'' = -3$

S = {2, -3}
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$c)\\ x^2 + 4x - 5 = 0$

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

a=1, b=4, c=−5
Δ=b2−4ac
Δ=(4)2−4*(1)*(−5)
Δ=16+20
Δ=36

$x = \frac{-4 \pm \sqrt{36}}{2*1}\\ \\x = \frac{-4 \pm 6}{2}\\ \\x' = \frac{-4 + 6}{2}\\ \\ x' = \frac{2}{2}\\ \\x' = 1\\ \\ \\x'' = \frac{-4 - 6}{2}\\ \\x'' = \frac{-10}{2}\\ \\x'' = -5$

S = {1, -5}
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$d) x^2-10x + 24 = 0$

a=1, b=−10, c=24

Δ=b2−4ac
Δ=(−10)2−4*(1)*(24)
Δ=100−96
Δ=4
$x = \frac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \frac{-(-10) \pm \sqrt{4}}{2*1}\\ \\x = \frac{10 \pm 2}{2}\\ \\ x' = \frac{10 + 2}{2}\\ \\x' = \frac{12}{2}\\ \\x' = 6\\ \\ \\ x'' = \frac{10 - 2}{2}\\ \\ x'' = \frac{8}{2}\\ \\x'' = 4$

S = {6, 4}
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$e)\\2x^2-9x+4 = 0$

a=2, b=−9, c=4
Δ=b2−4ac
Δ=(−9)2−4*(2)*(4)
Δ=81−32
Δ=49
$x = \frac{-b \pm \sqrt{\triangle}}{2*a}\\ \\x = \frac{-(-9) \pm \sqrt{49}}{2*2}\\ \\x = \frac{9 \pm 7}{4}\\ \\x' = \frac{9 + 7}{4}\\ \\x' = \frac{16}{4}\\ \\x' = 4\\ \\ \\x'' = \frac{9 + 7}{4}\\ \\x'' = \frac{2}{4}\\ \\x'' = \frac{1}{2}$


S = {4,$\frac{1}{2}$}
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$f)\\x^2+8x+16 = 0$

a=1, b=8, c=16
Δ=b2−4ac
Δ=(8)2−4*(1)*(16)
Δ=64−64
Δ=0
$x = \frac{-b \pm \sqrt{\triangle}}{2*a}\\ \\x = \frac{-8 \pm \sqrt{0}}{2*1}\\ \\x = \frac{-8 \pm 0}{2*1}\\ \\x' = \frac{-8 + 0}{2}\\ \\x ' = -4\\ \\x'' = \frac{-8 - 0}{2}\\ \\x''= -4$

S = {-4}

Answer 2

As respostas estão na imagem, se não entender algo pergunte-me. ;p