Question

amigos uma ajuda

1-coloque cada equação na sua forma normal ax+bx+c=0 e depois calcule a soma o produto e a soma do inverso das raizes:
a) 5x^2-11x-3=-5
b) 3x^2+7x+2=14x
c)-5x^2-7x-4=-7x^2
d) x^2+2/9=3
2-dados abaixo alguns pares de raizes mostre-me a equação do 2°grau:
a) 1e 6
b) 4 e -3
c) 4+√3 e 4- √3
d) 1/4 e 1/4
e) 2/5 e -1

Answer 1

Soma dos inversos das raízes (s):

$s=\dfrac{1}{x'}+\dfrac{1}{x''}\\\\\\s=\dfrac{x''}{x'x''}+\dfrac{x'}{x'x''}\\\\\\s=\dfrac{x''+x'}{x'x''}\\\\\\s=\dfrac{S}{P}\\\\\\s=\dfrac{\left(-\frac{b}{a}\right)}{\left(\frac{c}{a}\right)}\\\\\\\boxed{\boxed{s=-\dfrac{b}{c}}}$

Uma equação do segundo grau pode ser escrita em função de suas raízes:

$ax^{2}+bx+c=0$

Sejam x' e x'' as raízes da equação:

$ax^{2}+bx+c=(x-x')(x-x'')=0$
__________________________

1)
a)

$5x^{2}-11x-3=-5\\5x^{2}-11x-3+5=0\\5x^{2}-11x+2=0$

Soma das raízes:

$S=-\dfrac{b}{a}=-\dfrac{(-11)}{5}=\dfrac{11}{5}$

Produto das raízes:

$P=\dfrac{c}{a}=\dfrac{2}{5}$

Soma dos inversos das raízes:

$s=-\dfrac{b}{c}=-\dfrac{(-11)}{2}=\dfrac{11}{2}$

b)

$3x^{2}+7x+2=14x\\3x^{2}+7x-14x+2=0\\3x^{2}-7x+2=0\\\\\\S=-\dfrac{b}{a}=-\dfrac{(-7)}{3}=\dfrac{7}{3}\\\\\\P=\dfrac{c}{a}=\dfrac{2}{3}\\\\\\s=-\dfrac{b}{c}=-\dfrac{(-7)}{2}=\dfrac{7}{2}$

c)

$-5x^{2}-7x-4=-7x^{2}\\-5x^{2}-7x-4+7x^{2}=0\\2x^{2}-7x-4=0\\\\\\S=-\dfrac{b}{a}=-\dfrac{(-7)}{2}=\dfrac{7}{2}\\\\\\P=\dfrac{c}{a}=\dfrac{-4}{2}=-2\\\\\\s=-\dfrac{b}{c}=-\dfrac{(-7)}{(-4)}=-\dfrac{7}{4}$

d)

$\dfrac{x^{2}+2}{9}=3~~~~~~ou~~~~~~x^{2}+\dfrac{2}{9}=3~~~?$

2)
a)

$(x-1)(x-6)=0\\x^{2}-6x-x+6=0\\x^{2}-7x+6=0$

b)

$(x-4)(x-[-3])=0\\(x-4)(x+3)=0\\x^{2}+3x-4x-12=0\\x^{2}-x-12=0$

c)

$(x-[4+\sqrt{3}])(x-[4-\sqrt{3}])=0\\x^{2}-x(4-\sqrt{3})-x(4+\sqrt{3})+(4+\sqrt{3})(4-\sqrt{3})=0\\x^{2}-x[4-\sqrt{3}+4+\sqrt{3}]+(4^{2}-[\sqrt{3}]^{2})=0\\x^{2}-x(8)+(16-3)=0\\x^{2}-8x+13=0$

d)

$\left(x-\dfrac{1}{4}\right)\left(x-\dfrac{1}{4}\right)=0\\\\\\\left(\dfrac{4x-1}{4}\right)\left(\dfrac{4x-1}{4}\right)=0\\\\\\\left(\dfrac{4x-1}{4}\right)^{2}=0\\\\\\\dfrac{(4x-1)^{2}}{4^{2}}=0\\\\\\(4x-1)^{2}=0\\\\\\(4x)^{2}-2\cdot4x\cdot1+1^{2}=0\\\\\\16x^{2}-8x+1=0$

e)

$\left(x-\dfrac{2}{5}\right)(x-[-1])=0\\\\\\\left(\dfrac{5x-2}{5}\right)(x+1)=0\\\\\\\dfrac{(5x-2)(x+1)}{5}=0\\\\\\(5x-2)(x+1)=0\\\\\\5x^{2}+5x-2x-2=0\\\\\\5x^{2}+3x-2=0$