Question

URGENTE!!! PRECISO DESSA TAREFA PRA HOJE DIA 10/07/2020

Exercícios.

1) Resolva as equações do 2° grau.

A) x²-7x=0

B) x²+5x=0

C) 4x²-9x=0

D) 3x²+5x=0

E) 4x²-12x=0

F) 5x²+x=0


2) Como os coeficientes b e c é diferente de zero então essa é equação é completa.

A) x²-49=0

B) 3x²-7x+2=0

C) 2x²-7=15

D) x²=1

E) 2x²-50=0

F) 4x²+9=12x

G) 7x²-7=0

H) 5x²-15=0

i) 21=7x²

J) x²+9=4x

K) 25x²=20x-4

L) 5x²+20=0​

Answer 1

1)

$\mathtt{a)}~\mathsf{x^2-7x=0}\\\mathsf{x\cdot(x-7)=0}\\\mathsf{x=0~ou~x-7=0\to~x=7}\\\mathtt{b)}~\mathsf{x^2+5x=0}\\\mathsf{x\cdot(x+5)=0}\\\mathsf{x=0~ou~x+5=0\to~x=-5}\\\mathtt{c)}~\mathsf{4x^2-9x=0}\\\mathsf{x\cdot(4x-9)=0}\\\mathsf{x=0~ou~4x-9=0\to~x=\dfrac{9}{4}}\\\mathtt{d)}~\mathsf{3x^2+5x=0}\\\mathsf{x(3x+5)=0}\\\mathsf{x=0~ou~3x+5=0\to~x=-\dfrac{5}{3}}\\\mathtt{e)}~\mathsf{4x^2-12x=0}\\\mathsf{4x\cdot(x-3)=0}\\\mathsf{4x=0\to~x=0}\\\mathsf{x-3=0\to~x=3}\\\mathtt{f)}~\mathsf{5x^2+x=0}\\\mathsf{x\cdot(5x+1)=0}\\\mathsf{x=0~ou~5x+1=0\to~x=-\dfrac{1}{5}}$

$\dotfill$

2)

$\mathtt{a)}~\mathsf{x^2-49=0}\\\mathsf{x^2=49\to~x=\pm\sqrt{49}=\pm7}$

$\mathtt{b)}~\mathsf{3x^2-7x+2=0}\\\mathsf{a=3~~b=-7~~c=2}\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-7)^2-4\cdot3\cdot2}\\\mathsf{\Delta=49-24}\\\mathsf{\Delta=25}\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{x=\dfrac{-(-7)\pm\sqrt{25}}{2\cdot3}}\\\mathsf{x=\dfrac{7\pm5}{6}}\begin{cases}\mathsf{x_{1}=\dfrac{7+5}{6}=\dfrac{12}{6}=2}\\\mathsf{x_{2}=\dfrac{7-5}{6}=\dfrac{2\div2}{6\div2}=\dfrac{1}{3}}\end{cases}$

$\mathtt{c)}~\mathsf{2x^2-7=15}\\\mathsf{2x^2=22\to~x^2=11}\\\mathsf{x=\pm\sqrt{11}}$

$\mathtt{f)}~\mathsf{4x^2+9=12x}\\\mathsf{4x^2-12x+9=0}\\\mathsf{(2x-3)^2=0}\\\mathsf{2x-3=0}\\\mathsf{2x=3}\\\mathsf{x=\dfrac{3}{2}}$

$\mathtt{j)}~\mathsf{x^2+9=4x}\\\mathsf{x^2-4x+9=0}\\\mathsf{a=1~~b=-4~~c=9}\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-4)^2-4\cdot1\cdot9}\\\mathsf{\Delta=16-36=-20\implies~\not\exists~x\in\mathbb{R}}$

$\mathtt{k)}~\mathsf{25x^2=20x-4}\\\mathsf{25x^2-20x+4=0}\\\mathsf{(5x-2)^2=0}\\\mathsf{5x-2=0}\\\mathsf{5x=2}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{x=\dfrac{2}{5}}}}}}$

Answer 2

Resposta:

Explicação passo-a-passo:

1) Resolva as equações do 2° grau.

A) x²-7x=0

X.(x-7)= 0

x = 0

x-7= 0

x = 7

R.: V = {0;-7}

------------

B) x²+5x=0

x.(x+5)= 0

x = 0

x +5= 0

x = - 5

R.: V = {0; -5}

--------------

C) 4x²-9x=0

x.(4x-9)= 0

x = 0

4x-9= 0

4x = 9

x = 9/4

R.: v = {0; 9/4}

-------------

D) 3x²+5x=0

x.(3x+5)= 0

x = 0

3x+5= 0

3x = - 5

x = - 5/3

R.: {0; -5/3}

-----------------

E) 4x²-12x=0

4x.(x-3)= 0

4x = 0

x = 0/4

x = 0

x -3=0

x = 3

R.: V = {0; 3}

--------------

F) 5x²+x=0

x.(5x+1)= 0

x = 0

5x+1= 0

5x = - 1

x = -1/5

R.: V = {0; -1/5}

2) Como os coeficientes b e c é diferente de zero então essa é equação é completa.

A) x²-49=0 (incompleta)

x=√49

x = +/- 7

R.: {7; -7}

__________

B) 3x²-7x+2=0 (completa)

a = 3; b = -7; c = 2

∆= (-7)^2 - 4.3.2

∆= 49-24

∆= 25

√∆= 5

x = [ - b +/- √∆] 2a

x = [-(-7)+/- 5]/2.3

x =[7+/-5]/6

x' = (7+5)/6= 12/6= 2

x" = [7-5]/6 = 2/6 (:2)/(:2) = 1/3

V = {2; 1/3}

C) 2x²-7=15 (incompleta)

2x^2 = 15+7

2x^2 = 22

x^2= 22/2

x^2= 11

x = +/- √11

R : {11; -11}

_______

D) x²=1 (incompleta)

X =√1

X = +/- 1

R.: {1; -1}

_______

E) 2x²-50=0 (incompleta)

2x^2= 50

x^2 = 25

x = √25

x = +/- 5

R.: {5; -5}

___________

F) 4x²+9=12x (completa)

4x² - 12x + 9 = 0

a = 4; b = -12; c = 9

∆= b² - 4ac

∆= (-12)² - 4.4.9

∆= 144-144

∆= 0

x = [ - b +/- √∆]/2a

x = [ -(-12)+/- 0]/2.4

x = 12/2.4

x = 12/8 (:4)/(:4)

x = 3/2

R.: {3/2}

__________

G) 7x²-7=0 (completa)

7x^2 = 7

x^2 = 1

x = √1

x = +/- 1

R.: {1; -1}

__________

H) 5x²-15=0 (incompleta)

5x² = 15

x² = 15/5

x² = 3

x = +/- √3

R.: {√3; -√3}

___________

i) 21=7x² (incompleta)

7x² = 21

x² = 21/7

x = +/- √3

R.: {√3; -√3}

__________

J) x²+9=4x (completa)

x² - 4x + 9 = 0

a = 1; b = - 4; c = 9

∆= b² - 4ac

∆ = (-4)² - 4.1.9

∆= 16-36

∆= - 20

R.:

∆= - 20 (não há solução para os Reais)

∆<0

------------

K) 25x²=20x-4 (completa)

25x² - 20x+4= 0

a = 25; b = -20; c = 4

∆= b² - 4ac

∆= (-20)² - 4.25.4

∆= 400 - 400

∆= 0

x = [ 20+/-0]/2.25

x = 20/50

x = 2/5

R.: 2/5

L) 5x²+20=0 (incompleta)

5x² = - 20

x² = -20/5

x² = - 4

x = √(-4)

Não há solução para os Reais

(Não possui raiz quadrada de número negativo)