Question

1-Resolva as equações do 2° grau pela fórmula de Bhaskara, pelo método completando os quadrados e pela a soma e o produto das raízes.

1) x² - 10x + 21 = 0
2) -3x² - 5x + 2 = 0

Answer 1

Resposta:

Olá!

1)Báskara x² - 10x + 21 = 0

Δ = (-10)² - 4 . 1 . 21

Δ = 100 - 84

Δ = 16

x = - (-10) +- √16 / 2 . 1

x = 10 +- 4 / 2

x' = 14/2 = 7

x'' = 6/2 = 3

S {7, 3}

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2)Formulas: Δ = b² - 4.a.c

                x= -b ± √Δ / 2a

A=3 B=-5 C=2

Δ = b² - 4.a.c

Δ = (-5)² - 4.3.2

Δ = 25 - 24

Δ = 1

x= -b ± √Δ / 2a 

x= -(-5) ± √1 / 2.3

x= =5± 1 / 6

x' = 5+1/6

x' = 6/6 

x' = 1

x'' = 5 - 1/6

x'' = 4/6 (simplifica)

x'' = 2/3

A reposta será x' = 1 e x'' = 2/3

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Espero que lhe ajude bastante!!!!

Explicação passo-a-passo:

Answer 2

1)

pelo complemento de quadrados

$\mathsf{{x}^{2}-10x+21=0}\\\mathsf{{x}^{2}-10x=-21}\\\mathsf{{x}^{2}-10x+25=25-21}\\\mathsf{{(x-5)}^{2}=4}\\\mathsf{x-5=\pm\sqrt{4}}\\\mathsf{x-5=\pm2}$

$\mathsf{x-5=2\to~x=5+2=7}\\\mathsf{x-5=-2\to~x=5-2=3}$

$\boxed{\boxed{\mathsf{s=\{3,7\}}}}$

Pela "fórmula de Bháskara":

$\mathsf{{x}^{2}-10x+21=0}$

$\mathsf{\Delta={b}^{2}-4ac}\\\mathsf{\Delta={(-10)}^{2}-4.1.21}\\\mathsf{\Delta=100-84=16}$

$\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{x=\dfrac{-(-10)\pm\sqrt{16}}{2.1}}\\\mathsf{x=\dfrac{10\pm4}{2}}$

$\mathsf{x_{1}=\dfrac{10+4}{2}=\dfrac{14}{2}=7}\\\mathsf{x_{2}=\dfrac{10-4}{2}=\dfrac{6}{2}=3}$

$\boxed{\boxed{\mathsf{s=\{3,7\}}}}$

2)

Pelo complemento de quadrados

$\mathsf{-3{x}^{2}-5x+2=0\div(-3)}\\\mathsf{{x}^{2}+\dfrac{5}{3}x-\dfrac{2}{3}=0}\\\mathsf{{x}^{2}+\dfrac{5}{3}x=\dfrac{2}{3}}\\\mathsf{{x}^{2}+\dfrac{5}{3}x+\dfrac{25}{36}=\dfrac{25}{36}+\dfrac{2}{3}\times(36)}$

$\mathsf{36{x}^{2}+60x+25=25+24}\\\mathsf{{(6x+5)}^{2}=49}\\\mathsf{6x+5=\pm\sqrt{49}}\\\mathsf{6x+5=\pm7}$

$\mathsf{6x+5=7}\\\mathsf{6x=7-5}\\\mathsf{6x=2\to~x=\dfrac{2\div2}{6\div2}=\dfrac{1}{3}}$

$\mathsf{6x+5=-7}\\\mathsf{6x=-5-7}\\\mathsf{6x=-12\to~x=-\dfrac{12}{6}=-2}$

$\boxed{\boxed{\mathsf{s=\{-2,\dfrac{1}{3}\}}}}$

Pela "fórmula de Bháskara":

$\mathsf{-3{x}^{2}-5x+2=0\times(-1)}\\\mathsf{3{x}^{2}+5x-2=0}$

$\mathsf{\Delta={b}^{2}-4ac}\\\mathsf{\Delta={5}^{2}-4.3.(-2)}\\\mathsf{\Delta=25+24=49}$

$\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{x=\dfrac{-5\pm\sqrt{49}}{2.3}}\\\mathsf{x=\dfrac{-5\pm7}{6}}$

$\mathsf{x_{1}=\dfrac{-5-7}{6}=-\dfrac{12}{6}=-2}$

$\mathsf{x_{2}=\dfrac{-5+7}{6}=\dfrac{2}{6}=\dfrac{1}{3}}$

$\boxed{\boxed{\mathsf{s=\{-2,\dfrac{1}{3}\}}}}$