Question

Resolva cada equação abaixo, aplicando a ideia dos exemplos acima
pfv me ajudem

a) 13.x^2=100
b) 16.x^2-1225=0
c) 14.x^2=25
d) 17.10x^2-9=481
e) 15.2x^2-162=0
f) 18.x^2+13=94​

Answer 1

Explicação passo-a-passo:

a)

$\sf 13x^2=100$

$\sf x^2=\dfrac{100}{13}$

$\sf x=\pm\sqrt{\dfrac{100}{13}}$

• $\sf x'=\dfrac{10}{\sqrt{13}}$

$\sf x'=\dfrac{10}{\sqrt{13}}\cdot\dfrac{\sqrt{13}}{\sqrt{13}}$

$\sf \red{x'=\dfrac{10\sqrt{13}}{13}}$

• $\sf x"=\dfrac{-10}{\sqrt{13}}$

$\sf x"=\dfrac{-10}{\sqrt{13}}\cdot\dfrac{\sqrt{13}}{\sqrt{13}}$

$\sf \red{x"=\dfrac{-10\sqrt{13}}{13}}$

O conjunto solução é $\sf S=\Big\{\dfrac{-10\sqrt{13}}{13},\dfrac{10\sqrt{13}}{13}\Big\}$

b)

$\sf 16x^2-1225=0$

$\sf 16x^2=1225$

$\sf x^2=\dfrac{1225}{16}$

$\sf x=\pm\sqrt{\dfrac{1225}{16}}$

• $\sf \red{x'=\dfrac{35}{4}}$

• $\sf \red{x"=\dfrac{-35}{4}}$

O conjunto solução é $\sf S=\Big\{\dfrac{-35}{4},\dfrac{35}{4}\Big\}$

c)

$\sf 14x^2=25$

$\sf x^2=\dfrac{25}{14}$

$\sf x=\pm\sqrt{\dfrac{25}{14}}$

• $\sf x'=\dfrac{5}{\sqrt{14}}$

$\sf x'=\dfrac{5}{\sqrt{14}}\cdot\dfrac{\sqrt{14}}{\sqrt{14}}$

$\sf \red{x'=\dfrac{5\sqrt{14}}{14}}$

• $\sf x"=\dfrac{-5}{\sqrt{14}}$

$\sf x"=\dfrac{-5}{\sqrt{14}}\cdot\dfrac{\sqrt{14}}{\sqrt{14}}$

$\sf \red{x"=\dfrac{-5\sqrt{14}}{14}}$

O conjunto solução é $\sf S=\Big\{\dfrac{-5\sqrt{14}}{14},\dfrac{5\sqrt{14}}{14}\Big\}$

d)

$\sf 17\cdot10x^2-9=481$

$\sf 170x^2=481+9$

$\sf 170x^2=490$

$\sf x^2=\dfrac{490}{170}$

$\sf x^2=\dfrac{49}{17}$

$\sf x=\pm\sqrt{\dfrac{49}{17}}$

• $\sf x'=\dfrac{7}{\sqrt{17}}$

$\sf x'=\dfrac{7}{\sqrt{17}}\cdot\dfrac{\sqrt{17}}{\sqrt{17}}$

$\sf \red{x'=\dfrac{7\sqrt{17}}{17}}$

• $\sf x"=\dfrac{-7}{\sqrt{17}}$

$\sf x"=\dfrac{-7}{\sqrt{17}}\cdot\dfrac{\sqrt{17}}{\sqrt{17}}$

$\sf \red{x"=\dfrac{-7\sqrt{17}}{17}}$

O conjunto solução é $\sf S=\Big\{\dfrac{-7\sqrt{17}}{17},\dfrac{7\sqrt{17}}{17}\Big\}$

e)

$\sf 15\cdot2x^2-162=0$

$\sf 30x^2=162$

$\sf x^2=\dfrac{162}{30}$

$\sf x^2=\dfrac{81}{15}$

$\sf x=\pm\sqrt{\dfrac{81}{15}}$

• $\sf x'=\dfrac{9}{\sqrt{15}}$

$\sf x'=\dfrac{9}{\sqrt{15}}\cdot\dfrac{\sqrt{15}}{\sqrt{15}}$

$\sf x'=\dfrac{9\sqrt{15}}{15}$

$\sf \red{x'=\dfrac{3\sqrt{15}}{5}}$

• $\sf x"=\dfrac{-9}{\sqrt{15}}$

$\sf x"=\dfrac{-9}{\sqrt{15}}\cdot\dfrac{\sqrt{15}}{\sqrt{15}}$

$\sf x"=\dfrac{-9\sqrt{15}}{15}$

$\sf \red{x"=\dfrac{-3\sqrt{15}}{5}}$

O conjunto solução é $\sf S=\Big\{\dfrac{-3\sqrt{15}}{5},\dfrac{3\sqrt{15}}{5}\Big\}$

f)

$\sf 18x^2+13=94$

$\sf 18x^2=94-13$

$\sf 18x^2=81$

$\sf x^2=\dfrac{81}{18}$

$\sf x^2=\dfrac{9}{2}$

$\sf x=\pm\sqrt{\dfrac{9}{2}}$

• $\sf x'=\dfrac{3}{\sqrt{2}}$

$\sf x'=\dfrac{3}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf \red{x'=\dfrac{3\sqrt{2}}{2}}$

• $\sf x"=\dfrac{-3}{\sqrt{2}}$

$\sf x"=\dfrac{-3}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf \red{x"=\dfrac{-3\sqrt{2}}{2}}$

O conjunto solução é $\sf S=\Big\{\dfrac{-3\sqrt{2}}{2},\dfrac{3\sqrt{2}}{2}\Big\}$