Question

usando o processo geométrico al-Khwarizmi,determine as raizes de cada uma das seguintes equações do 2º grau com uma incógnita no conjunto dos números reais:

EU SÓ QUERO OS ITENS “K” E “L”
PF ME AJUDEM

Answer 1

k)

$\mathsf{3{x}^{2}-2x-1=0\div3}\\\mathsf{{x}^{2}-\dfrac{2}{3}x-1=0}\\\mathsf{{x}^{2}-\dfrac{2}{3}x=1}\\\mathsf{{x}^{2}+\dfrac{2}{3}x+\dfrac{4}{9}=\dfrac{4}{9}+1}$

$\mathsf{9{x}^{2}+6x+4=4+9}\\\mathsf{{(3x+2)}^{2}=13}\\\mathsf{3x+2=\pm\sqrt{13}}}\\\mathsf{3x+2=\sqrt{13}}\\\mathsf{3x=\sqrt{13}-2}\\\mathsf{x=\dfrac{\sqrt{13}-2}{3}}$

$\mathsf{3x+2=-\sqrt{13}}\\\mathsf{3x=-\sqrt{13}-2}\\\mathsf{x=-\dfrac{\sqrt{13}+2}{3}}$

e)

$\mathsf{{x}^{2}+3x-10=0}\\\mathsf{{x}^{2}+3x=10}\\\mathsf{{x}^{2}+3x+\dfrac{9}{4}=10+\dfrac{9}{4}\times(4)}\\\mathsf{4{x}^{2}+12x+9=40+9}\\\mathsf{{(2x+3)}^{2}=49}\\\mathsf{2x+3=\pm\sqrt{49}}\\\mathsf{2x+3=\pm7}$

$\mathsf{2x+3=7}\\\mathsf{2x=7-3}\\\mathsf{2x=4\to~x=\dfrac{4}{2}=2}\\\mathsf{2x+3=-7}\\\mathsf{2x=-3-7}\\\mathsf{2x=-10\to~x=-\dfrac{10}{2}=-5}$

L)

$\mathsf{10{x}^{2}+7x+1=0\div10}\\\mathsf{{x}^{2}+\dfrac{7}{10}x+\dfrac{1}{10}=0}\\\mathsf{{x}^{2}+\dfrac{7}{10}x=-\dfrac{1}{10}}\\\mathsf{{x}^{2}+\dfrac{7}{10}x+\dfrac{49}{400}=\dfrac{49}{400}-\dfrac{1}{10}\times(400)}$

$\mathsf{400{x}^{2}+280x+49=49-40}\\\mathsf{{(20x+7)}^{2}=9}\\\mathsf{20x+7=\pm\sqrt{49}}\\\mathsf{20x+7=\pm7}\\\mathsf{20x+7=7}\\\mathsf{20x=7-7}\\\mathsf{20x=0\to\,x=\dfrac{0}{20}=0}$

$\mathsf{20x+7=-7}\\\mathsf{20x=-7-7}\\\maths{20x=-14\to~x=\dfrac{-14\div2}{20\div2}=-\dfrac{7}{10}}$

Answer 2

Resposta:

Explicação passo-a-passo:

k) 3x²-2x-1=0

3x²-2x=1      (÷3)

x²-2x/3=1/3

Lembre-se:

(x-a)²=x²-2ax+a²

-2ax=-2x/3 => a= 1/3

Voltando a equação:

x²-2x/3+(1/3)²=1/3+(1/3)²

(x-1/3)²=1/3+1/9

(x-1/3)²=(3+1)/9

(x-1/3)²=4/9

(x-1/3)=±√(4/9)

(x-1/3)=±2/3

1a solução:

(x-1/3)=2/3

x=2/3+1/3

x=1

2a solução:

(x-1/3)=-2/3

x=1/3-2/3

x= -1/3

S={1, -1/3}

l)

10x²+7x+1=0

10x²+7x= -1      (÷10)

x²+7x/10= -1/10

Lembre-se:

(x+a)²=x²+2ax+a²

2ax=7x/10

a=7/20

Voltando a equação:

x²+7x/10+(7/20)²=(7/20)²-1/10

(x+7/20)²=49/400-1/10

(x+7/20)²=(49-40)/400

(x+7/20)²=9/400

(x+7/20)=±√9/400

(x+7/20)=±3/20

1a solução:

x+7/20=3/20

x=3/20-7/20

x= -4/20= -1/5

2a solução:

x+7/20= -3/20

x= -3/20-7/20

x= -10/20

x= -1/2

S={-1/2, -1/5}