Question

1-) Racionalize:
a)10/√3
b)1/√23
c)8/√8
d)5/³√7²
e)3/^5√2²

2-) Determine o valor do discriminante ∆ e diga se a equação tem raízes reais: ∆=b²-4.a.c
a)2x²-3x+1=0
b)-3x²+10x-3=0
c)2x²+3x+25=0

3-) Encontre a soma e o produto das raízes, sem resolver q equação:
s= -b/2.a p=-∆/4.a
a)4x²+x-14=0
b)-9x²+x-8=0
c)x²+4x-5=0

4-) Determine os valores dos coeficientes a,b,c nas equações seguintes:
a)x²-7x+10=0
b)x²-x-6=0
c)4x²-4x+1=0
d)-x²-5x-1=0
e)4x²=0
f)3x²-7=0
g)x²-2x=0
h)12x²+4-5x=0

5-) Resolva as equações na forma ax²+bx=0 do 2º grau:
a)x²-8x=0
b)x²+3x=0
c)9x²-72=0
d)5x²-20=0

6-) A figura mostra um edifício que tem 15 metros de altura, com uma escada colocada a 8 metros de sua base ligada ao topo do edifício. o comprimento dessa escada é de:
a)12 m
b)30 m
c)15 m
d)17 m
e)20 m

Eu preciso disso urgente


​

Answer 1

Questão 1)

A racionalização consiste em tranformar a fração com denominador irracional em racional através da multiplicação por um radical

Letra A)

$\begin{array}{l}\sf \dfrac{10}{\sqrt{3}}=\\\\\sf =\dfrac{10}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}\\\\\!\boxed{\sf =\dfrac{10\sqrt{3}}{3}} \\ \\ \end{array}$

Letra B)

$\begin{array}{l}\sf \dfrac{1}{\sqrt{23}}=\\\\\sf =\dfrac{1}{\sqrt{23}}\cdot\dfrac{\sqrt{23}}{\sqrt{23}}\\\\\boxed{\sf =\dfrac{\sqrt{23}}{23}}\\\\\end{array}$

Letra C)

$\begin{array}{l}\sf \dfrac{8}{\sqrt{8}}=\\\\\sf =\dfrac{8}{2\sqrt{2}}\\\\\sf =\dfrac{4}{\sqrt{2}}\\\\\sf =\dfrac{4}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf =\dfrac{4\sqrt{2}}{2}\\\\\!\boxed{\sf =2\sqrt{2}}\\\\\end{array}$

Letra D)

$\begin{array}{l}\sf \dfrac{5}{\sqrt[\sf3]{\sf7^2}}=\\\\\sf =\dfrac{5}{\sqrt[\sf3]{\sf7^2}}\cdot\dfrac{\sqrt[\sf3]{\sf7}}{\sqrt[\sf3]{\sf7}}\\\\\sf =\dfrac{5\sqrt[\sf3]{\sf7}}{\sqrt[\sf3]{\sf7^3}}\\\\\!\boxed{\sf =\dfrac{5\sqrt[\sf3]{\sf7}}{7}}\\\\\end{array}$

Letra E)

$\begin{array}{l}\sf \dfrac{3}{\sqrt[\sf5]{2^2}}=\\\\\sf =\dfrac{3}{\sqrt[\sf5]{2^2}}\cdot\dfrac{\sqrt[\sf5]{2^3}}{\sqrt[\sf5]{2^3}}\\\\\sf = \dfrac{3\sqrt[\sf5]{2^3}}{\sf\sqrt[\sf5]{2^5}}\\\\\!\boxed{\sf = \dfrac{3\sqrt[\sf5]{8}}{2}}\\\\\end{array}$

ㅤ

Questão 2)

Para determinar se a equação terá raízes reais a partir do valor do delta, veja as regras a seguir

$\small\begin{array}{l}~~~\bullet~\sf Se~\:\Delta > 0~\to~x'~e~\:x''\!\in\mathbb{R},~\:com\:~x'\,\neq\, x''\\\\~~~\bullet~\sf Se~\:\Delta = 0~\to~x'~e~\:x''\!\in\mathbb{R},~\:com\:~x'= x''\\\\~~~\bullet~\sf Se~\:\Delta < 0~\to~x'~e~\:x''\!\notin\mathbb{R}\end{array}$

Assim:

Letra A)

$\begin{array}{l}\sf 2x^2-3x+1=0\\\\\sf \Delta=b^2-4ac\\\\\sf \Delta=(-3)^2-4(2)(1)\\\\\sf \Delta=9-8\\\\\!\boxed{\sf \Delta=1}\sf~\to~\Delta > 0\end{array}$

Assim a equação terá raízes reais

ㅤ

Letra B)

$\begin{array}{l}\sf -3x^2+10x-3=0\\\\\sf \Delta=b^2-4ac\\\\\sf \Delta=(10)^2-4(-3)(-3)\\\\\sf \Delta=100-36\\\\\!\boxed{\sf \Delta=64}\sf~\to~\Delta > 0\end{array}$

Assim a equação terá raízes reais

ㅤ

Letra C)

$\begin{array}{l}\sf 2x^2+3x+25=0\\\\\sf \Delta=b^2-4ac\\\\\sf \Delta=(3)^2-4(2)(25)\\\\\sf \Delta=9-200\\\\\!\boxed{\sf \Delta=-191}\sf~\to~\Delta < 0~\therefore~x\notin\mathbb{R}\end{array}$

Assim a equação não terá raízes reais

ㅤ

Questão 3)

Usando as fórmulas de soma e produto:

Letra A)

$\begin{array}{l}\sf 4x^2+x-14=0\\\\\sf S=-\dfrac{b}{a}\quad e\quad P=\dfrac{c}{a}\\\\\sf S=-\dfrac{1}{4}\quad e\quad P=\dfrac{-14~}{4}\\\\\!\boxed{\sf S=-\dfrac{1}{4}\quad e\quad P=-\dfrac{7}{2}} \\ \\ \end{array}$

Letra B)

$\begin{array}{l}\sf -9x^2+x-8=0\\\\\sf S=-\dfrac{b}{a}\quad e\quad P=\dfrac{c}{a}\\\\\sf S=-\dfrac{1}{-9~}\quad e\quad P=\dfrac{-8~}{-9~}\\\\\!\boxed{\sf S=\dfrac{1}{9}\quad e\quad P=\dfrac{8}{9}}\\\\\end{array}$

Letra C)

$\begin{array}{l}\sf x^2+4x-5=0\\\\\sf S=-\dfrac{b}{a}\quad e\quad P=\dfrac{c}{a}\\\\\sf S=-\dfrac{4}{1}\quad e\quad P=\dfrac{-5~}{1}\\\\\!\boxed{\sf S=-4\quad e\quad P=-5}\end{array}$

ㅤ

Questão 4)

Para identificar os coeficientes, lembre-se da forma geral:

ax² + bx + c = 0

ㅤ

Letra A)

x² – 7x + 10 = 0

a = 1, b = – 7, c = 10

ㅤ

Letra B)

x²– x – 6 = 0

a = 1, b = – 1, c = – 6

ㅤ

Letra C)

4x² – 4x + 1 = 0

a = 4, b = – 4, c = 1

ㅤ

Letra D)

– x² – 5x – 1 = 0

a = – 1, b = – 5, c = – 1

ㅤ

Letra E)

4x² = 0

a = 4, b = 0, c = 0

ㅤ

Letra F)

3x² – 7 = 0

a = 3, b = 0, c = – 7

ㅤ

Letra G)

x²– 2x = 0

a = 1, b = – 2, c = 0

ㅤ

Letra H)

12x² + 4 – 5x = 0

a = 12, b = – 5, c = 4

ㅤ

Questão 5)

Resolvendo as equações incompletas:

Letra A)

$\begin{array}{l}\sf x^2-8x=0\\\\\sf x(x-8)=0\\\\ \begin{cases}\sf x=0\\\\\sf x-8=0\end{cases}\\\\\begin{cases}\sf x'=0\\\\\sf x''=8\end{cases}\\\\\sf S=\Big\{0~~;~~8\Big\} \\ \\ \end{array}$

Letra B)

$\begin{array}{l}\sf x^2+3x=0\\\\\sf x(x+3)=0\\\\ \begin{cases}\sf x=0\\\\\sf x+3=0\end{cases}\\\\\begin{cases}\sf x'=0\\\\\sf x''=-3\end{cases}\\\\\sf S=\Big\{-3~~;~~0\Big\}\\\\\end{array}$

Letra C)

$\begin{array}{l}\sf 9x^2-72=0\\\\\sf 72+9x^2-72=0+72\\\\\sf 9x^2=72\\\\\sf \dfrac{9x^2}{9}=\dfrac{72}{9}\\\\\sf x^2=8\\\\\sf \sqrt{x^2}=\sqrt{8}\\\\\sf |x|=\sqrt{8}\\\\\sf x=\pm~\sqrt{8}\\\\\sf x=\pm~2\sqrt{2}\\\\\sf S=\Big\{-2\sqrt{2}~~;~~2\sqrt{2}\Big\}\\\\\end{array}$

Letra D)

$\begin{array}{l}\sf 5x^2-20=0\\\\\sf 20+5x^2-20=0+20\\\\\sf 5x^2=20\\\\\sf \dfrac{5x^2}{5}=\dfrac{20}{5}\\\\\sf x^2=4\\\\\sf \sqrt{x^2}=\sqrt{4}\\\\\sf |x|=2\\\\\sf x=\pm~2\\\\\sf S=\Big\{-2~~;~~2\Big\}\\\\\end{array}$

ㅤ

Questão 6)

Se imaginarmos, vemos um triângulo retângulo, onde temos que os catetos que são 15 m e 8 m, e a hipotenusa que é a altura da escada ( chamaremos de " h " )

Dessa forma aplicando o Teorema de Pitágoras:

$\begin{array}{l}\sf a^2=b^2+c^2\\\\\sf (h)^2=(15)^2+(8)^2\\\\\sf h^2=225+64\\\\\sf h^2=289\\\\\sf \sqrt{h^2}=\sqrt{289}\\\\\!\boxed{\sf h=17} \\ \\ \end{array}$

Assim, resposta é LETRA D) 17 m

ㅤ

Att. Nasgovaskov