Question

Me ajudem por favor eu suplico ​

Answer 1

Resposta:

÷

Explicação passo a passo:

$1) x^{2} +3x - 5 = 0\\a= 1\\b= 3\\c= -5\\\\-5x^{2} +2x = 0\\a= -5\\b= 2\\c=0\\\\2-4x^{2} =0\\a= -4\\b=0\\c=2$

$2) 5x^{2} +4x = -3\\5x^{2} +4x +3 = 0\\\\-x^{2} +3+6x=3x^{2} +1\\-x^{2} +3+6x -3x^{2} -1=0\\-4x^{2} +6x+2=0\\\\3)11x^{2} =0\\x^{2} =\frac{0}{11} \\x^{2} = 0\\x = \sqrt{0}\\ x = 0\\\\\frac{2}{9} x^{2} = 0\\2x^{2} = 0 *9\\2x^{2} = 0\\x^{2} = \frac{0}{2}\\x^{2} = 0\\x = \sqrt{0}\\ x = 0\\\\x^{2} -36 = 0\\x^{2} = 36\\x= \sqrt{36}\\ x= 6\\\\2x^{2} -8 = 0\\2x^{2} = 8\\x^{2} = \frac{8}{2}\\x^{2} = 4\\x = \sqrt{4}\\ x= 2$

$4x^{2} + 9x = 0\\4x^{2} = -9x\\x(4x) = x(-9)\\4x = -9\\x = \frac{-9}{4}$

$4) 3x^{2} - 7x + 4 = 0\\b^{2} - 4ac\\(-7)^{2} - 4*3*4 = 49 - 48 = 1\\\\x = \frac{ -b +- \sqrt{delta}}{2a}\\x = \frac{-(-7) +- \sqrt{1}}{2*3}\\x = \frac{7 +- 1}{6}\\x' = \frac{7 +1}{6} = \frac{8}{6} = \frac{4}{3}\\x'' = \frac{7 -1}{6} = \frac{6}{6} = 1$

$4x^{2} -12x+4 = 0 (dividido por 4 para facilitar)\\x^{2} -3x+1 = 0\\delta = (-3)^{2} - 4*1*1 = 9-4 = 5\\x = \frac{ -b +- \sqrt{delta}}{2a}\\x = \frac{-(-3) +- \sqrt{5}}{2*1}\\x = \frac{3 +- \sqrt{5}}{2}\\x' = \frac{3 +\sqrt{5}}{2}\\x'' = \frac{3- \sqrt{5}}{2}$

$x^{2} -5x + 6 = 0\\delta = (-5)^{2} - 4*1*6 = 25 - 24 = 1\\x = \frac{-(-5)+-\sqrt{1} }{2*1} = \frac{5+-{1} }{2}\\x' = \frac{5+1}{2} = \frac{6}{2} = 3\\x'' = \frac{5-1}{2} = \frac{4}{2} = 2$