Question

Resolva essas equações do 2º grau (podendo escolher o método; fatoração,Bkáskara ou algebricamente)
x² - 13x + 36 = 0
x² + 5x = 6
x² - 4x - 60 = 0
4x² + 12x - 7 = 0

Answer 1

Resoluções, veja:

Letra ''A'':

$x^{2}-13x+36=0\\\\\ x = -b~\pm\dfrac{ \sqrt{b^{2}-4~.~a~.~c}}{2~.~a}}\\\\\ x = 13~\pm\dfrac{ \sqrt{(-13)^{2}-4~.~1~.~36}}{2~.~1}}}\\\\\ x = 13~\pm\dfrac{ \sqrt{169-144}}{2}}\\\\\ x = 13~\pm\dfrac{ \sqrt{25}}{2}}\\\\\ x = 13~\pm\dfrac{5}{2}}\\\\ x' = \dfrac{13+5}{2}}\\\\\ x' = \dfrac{18}{2}~\therefore~\large\boxed{\boxed{x' = 9.}}\\\\\\\ x'' = \dfrac{13-5}{2}\\\\ x''=\dfrac{8}{2}~ \therefore~\large\boxed{\boxed{x'' = 4.}}}\\\\\\\ \large\boxed{\boxed{x' = 9~e~x''=4.}}$

Letra ''B'':

$x^{2}+5x=6 =\ \textgreater \ ~ x^{2}+5x-6=0\\\\\ x = -b~\pm\dfrac{ \sqrt{b^{2}-4~.~a~.~c}}{2~.~a}}\\\\\\ x = -5~\pm\dfrac{ \sqrt{5^{2}-4~.~1~.~(-6)}}{2~.~1}}\\\\\\ x = -5~\pm\dfrac{ \sqrt{25+24}}{2}}\\\\\ x = -5~\pm\dfrac{ \sqrt{49}}{2}}\\\\\ x = -5~\pm\dfrac{7}{2}\\\\\\ x' = \dfrac{-5+7}{2}\\\\\\ x' = \dfrac{2}{2}~\therefore~\large\boxed{\boxed{x' = 1.}}\\\\\\ x'' = \dfrac{-5-7}{2}\\\\\\ x'' = \dfrac{-12}{2}~\therefore~\large\boxed{\boxed{x''=-6.}}\\\\\\\\ \large\boxed{\boxed{x'=1~e~x''=-6.}}$

Letra ''C'':

$x^{2}-4x-60=0\\\\\ x = -b~\pm\dfrac{ \sqrt{b^{2}-4~.~a~.~c}}{2~.~a}}\\\\\ x = 4~\pm\dfrac{ \sqrt{(-4)^{2}-4~.~1~.~(-60)}}{2~.~1}}\\\\\ x = 4~\pm\dfrac{ \sqrt{16+240}}{2}}\\\\\ x =4~\pm\dfrac{ \sqrt{256}}{2}}\\\\\ x =4~\pm\dfrac{16}{2}}\\\\\\ x'= \dfrac{4+16}{2}}\\ \\\\ x' = \dfrac{20}{2}}~\therefore~\large\boxed{\boxed{x'=10.}}\\\\\\ x'' = \dfrac{4-16}{2}\\\\\\\ x'' = \dfrac{-12}{2}}~\therefore~\large\boxed{\boxed{x''=-6.}}}\\\\\\\\ \large\boxed{\boxed{x'=10~e~x''=-6.}}$

Letra ''D'':

$4x^{2}+12x-7=0\\\\\ x = -b~\pm\dfrac{ \sqrt{b^{2}-4~.~a~.~c}}{2~.~a}}\\\\\ x = -12~\pm\dfrac{ \sqrt{12^{2}-4~.~4~.~(-7)}}{2~.~4}}\\\\\ x = -12~\pm\dfrac{ \sqrt{144+112}}{8}\\\\\ x = -12~\pm\dfrac{ \sqrt{256}}{8}\\\\\ x = -12~\pm\dfrac{16}{8}\\\\\\\ x' = \dfrac{-12 +16}{8}\\\\\\\ x' = \dfrac{4}{8}~\therefore~\large\boxed{\boxed{x'= \dfrac{1}{2}.}}\\\\\\\\ x''= \dfrac{-12-16}{8}}\\\\\\ x'' = \dfrac{-28}{8}}~\therefore~\large\boxed{\boxed{x''=\dfrac{-7}{2}.}}$

Espero que te ajude :).