Resolucao_da_lista_equacoes_completas_de_2_grau

Prévia do material em texto

1 
Cálculo: conceitos – Aula 2 
 Resolução da lista de exercícios de revisão equação de 2º grau 
 
 
1) Resolva as seguintes equações de 2º Grau, utilizando a fórmula de Bhaskara. 
 
a) x² + 7x – 60 = 0 
𝒂 = 𝟏; 𝒃 = 𝟕; 𝒄 = −𝟔𝟎 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
−𝟕 ± √𝟒𝟗 − 𝟒. 𝟏. (−𝟔𝟎)
𝟐
=
−𝟕 ± √𝟒𝟗 + 𝟐𝟒𝟎
𝟐
=
−𝟕 ± √𝟐𝟖𝟗
𝟐
 
𝒙 =
−𝟕 ± 𝟏𝟕
𝟐
→ {
𝒙′ =
−𝟕 + 𝟏𝟕
𝟐
=
𝟏𝟎
𝟐
= 𝟓
𝒙′′ =
−𝟕 − 𝟏𝟕
𝟐
=
−𝟐𝟒
𝟐
= −𝟏𝟐
↔ 𝑺 = {−𝟏𝟐, 𝟓} 
b) x² + 4x - 21 = 0 
𝒂 = 𝟏; 𝒃 = 𝟒; 𝒄 = −𝟐𝟏 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
−𝟒 ± √𝟏𝟔 − 𝟒. 𝟏. (−𝟐𝟏)
𝟐
=
−𝟒 ± √𝟏𝟔 + 𝟖𝟒
𝟐
=
−𝟒 ± √𝟏𝟎𝟎
𝟐
 
𝒙 =
−𝟒 ± 𝟏𝟎
𝟐
→ {
𝒙′ =
−𝟒 + 𝟏𝟎
𝟐
=
𝟔
𝟐
= 𝟑
𝒙′′ =
−𝟒 − 𝟏𝟎
𝟐
=
−𝟏𝟒
𝟐
= −𝟕
↔ 𝑺 = {−𝟕, 𝟑} 
c) 2x² + 4x + 8 = 0 
𝒂 = 𝟐; 𝒃 = 𝟒; 𝒄 = 𝟖 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
−𝟒 ± √𝟏𝟔 − 𝟒. 𝟐. 𝟖
𝟐. 𝟐
=
−𝟒 ± √𝟏𝟔 − 𝟔𝟒
𝟒
=
−𝟒 ± √−𝟒𝟖
𝟒
 
√−𝟒𝟖 ∉ ℝ ↔ 𝑺 = ∅ 
 
d) y² - 11y + 18 = 0 
𝒂 = 𝟏; 𝒃 = −𝟏𝟏; 𝒄 = 𝟏𝟖 
𝒚 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
𝟏𝟏 ± √𝟏𝟐𝟏 − 𝟒. 𝟏. 𝟏𝟖
𝟐
=
𝟏𝟏 ± √𝟏𝟐𝟏 − 𝟕𝟐
𝟐
=
𝟏𝟏 ± √𝟒𝟗
𝟐
 
𝒚 =
𝟏𝟏 ± 𝟕
𝟐
→ {
𝒚′ =
𝟏𝟏 + 𝟕
𝟐
=
𝟏𝟖
𝟐
= 𝟗
𝒚′′ =
𝟏𝟏 − 𝟕
𝟐
=
𝟒
𝟐
= 𝟐
↔ 𝑺 = {𝟐, 𝟗} 
e) -3x² + 5x = 0 
𝒂 = −𝟑; 𝒃 = 𝟓; 𝒄 = 𝟎 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
−𝟓 ± √𝟐𝟓 − 𝟒. (−𝟑). 𝟎
𝟐. (−𝟑)
=
−𝟓 ± √𝟐𝟓
−𝟔
=
−𝟓 ± 𝟓
−𝟔
 
𝒙 =
−𝟓 ± 𝟓
−𝟔
→ {
𝒙′ =
−𝟓 + 𝟓
−𝟔
=
𝟎
−𝟔
= 𝟎
𝒙′′ =
−𝟓 − 𝟓
−𝟔
=
−𝟏𝟎
−𝟔
=
𝟏𝟎
𝟔
=
𝟓
𝟑
↔ 𝑺 = {𝟎,
𝟓
𝟑
} 
 
 
 
 
2 
 
f) 2y² + 4y - 6 = 0 
𝒂 = 𝟐; 𝒃 = 𝟒; 𝒄 = −𝟔 
𝒚 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
−𝟒 ± √𝟏𝟔 − 𝟒. 𝟐. (−𝟔)
𝟒
=
−𝟒 ± √𝟏𝟔 + 𝟒𝟖
𝟒
=
−𝟒 ± √𝟔𝟒
𝟒
 
𝒚 =
−𝟒 ± 𝟖
𝟒
→ {
𝒚′ =
−𝟒 + 𝟖
𝟒
=
𝟒
𝟒
= 𝟏
𝒚′′ =
−𝟒 − 𝟖
𝟒
=
−𝟏𝟐
𝟒
= −𝟑
↔ 𝑺 = {−𝟑, 𝟏} 
g) x² - 10x +25 = 0 
𝒂 = 𝟏; 𝒃 = −𝟏𝟎; 𝒄 = 𝟐𝟓 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
𝟏𝟎 ± √𝟏𝟎𝟎 − 𝟒. 𝟏. 𝟐𝟓
𝟐
=
𝟏𝟎 ± √𝟏𝟎𝟎 − 𝟏𝟎𝟎
𝟐
=
𝟏𝟎 ± √𝟎
𝟐
 
𝒙 =
𝟏𝟎 ± 𝟎
𝟐
→ {
𝒙′ =
𝟏𝟎 + 𝟎
𝟐
=
𝟏𝟎
𝟐
= 𝟓
𝒙′′ =
𝟏𝟎 − 𝟎
𝟐
=
𝟏𝟎
𝟐
= 𝟓
↔ 𝑺 = {𝟓} 
h) x² - x - 6 = 0 
𝒂 = 𝟏; 𝒃 = −𝟏; 𝒄 = −𝟔 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
𝟏 ± √𝟏 − 𝟒. 𝟏. (−𝟔)
𝟐
=
𝟏 ± √𝟏 + 𝟐𝟒
𝟐
=
𝟏 ± √𝟐𝟓
𝟐
 
𝒙 =
𝟏 ± 𝟓
𝟐
→ {
𝒙′ =
𝟏 + 𝟓
𝟐
=
𝟔
𝟐
= 𝟑
𝒙′′ =
𝟏 − 𝟓
𝟐
=
−𝟒
𝟐
= −𝟐
↔ 𝑺 = {−𝟐, 𝟑} 
i) 2x² + 4x - 70 = 0 
𝒂 = 𝟐; 𝒃 = 𝟒; 𝒄 = −𝟕𝟎 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
−𝟒 ± √𝟏𝟔 − 𝟒. 𝟐. (−𝟕𝟎)
𝟐. 𝟐
=
−𝟒 ± √𝟏𝟔 + 𝟓𝟔𝟎
𝟒
=
−𝟒 ± √𝟓𝟕𝟔
𝟒
 
𝒙 =
−𝟒 ± 𝟐𝟒
𝟒
→ {
𝒙′ =
−𝟒 + 𝟐𝟒
𝟒
=
𝟐𝟎
𝟒
= 𝟓
𝒙′′ =
−𝟒 − 𝟐𝟒
𝟒
=
−𝟐𝟖
𝟒
= −𝟕
↔ 𝑺 = {−𝟕, 𝟓} 
j) x² -18x + 81 = 0 
𝒙 = 𝟏; 𝒃 = −𝟏𝟖; 𝒄 = 𝟖𝟏 
𝒙 =
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
=
𝟏𝟖 ± √𝟑𝟐𝟒 − 𝟒. 𝟏. 𝟖𝟏
𝟐
=
𝟏𝟖 ± √𝟑𝟐𝟒 − 𝟑𝟐𝟒
𝟐
=
𝟏𝟖 ± √𝟎
𝟐
 
𝒙 =
𝟏𝟖 ± 𝟎
𝟐
→ {
𝒙′ =
𝟏𝟖 + 𝟎
𝟐
=
𝟏𝟖
𝟐
= 𝟗
𝒙′′ =
𝟏𝟖 − 𝟎
𝟐
=
𝟏𝟖
𝟐
= 𝟗
↔ 𝑺 = {𝟗}