Question

Resolva as equações de 2° grau : (a) 3x ao quadrado - 4 x 2 = 0. (B) 36 = 4 y ao quadrado. (C) - x ao quadrado - x + 30 = 0. (D) n ( n + 3 ) - 40 = 0. (E) ( y + 2 ) ( y - 3 ) = 0. (F) ( 2 x - 4 ) ao quadrado = 0. Me Ajudem

Answer 1

a) $3x^{2} - x4^{2} = 0 \\ -x^{2} =0==\ \textgreater \ +/-x=0$

b) $\sqrt{ 4y^{2} } = \sqrt{36} \\ 2y=6==\ \textgreater \ y= \frac{6}{2} ==\ \textgreater \ y=3$

c) $x' = \frac{1+\sqrt{1-(-4)*1*30} }{2} ==\ \textgreater \ x'= 1+ \sqrt{121} ==\ \textgreater \ x'=12 \\ x" = \frac{1-\sqrt{1-(-4)*1*30} }{2} ==\ \textgreater \ x"= 1- \sqrt{121} ==\ \textgreater \ x" = (-10)$

d)  $n^{2} +3n-40=0 \\ x'= \frac{-1+ \sqrt{9-4*1*(-40)} }{2} = \frac{-1+13}{2} ==\ \textgreater \ x'=6 \\ x"=\frac{-1-13}{2} ==\ \textgreater \ x"=\frac{-1-13}{2} ==\ \textgreater \ x"=-7$

e) $y^{2} -y-6=0 \\ x'= \frac{1+ \sqrt{1-4*1*(-6)} }{2} = \frac{1+ \sqrt{25} }{2}= \frac{1+5}{2} ==\ \textgreater \ x'=3 \\ x"=\frac{1-5}{2} ==\ \textgreater \ x"=-2$

f) $4 x^{2} -16x+16=0 ==\ \textgreater \ x^{2} -4x+4=0 \\ x'= \frac{4+ \sqrt{ 4^{2} -4*1*4} }{2} = \frac{4+ \sqrt{0} }{2} ==\ \textgreater \ x'=x"= \frac{4}{2} =2$