Question

me ajudem pfvr

aplicando a forma de Bhaskara determine o valor de x (a)-2x²+3+5=0
(b) x²-5x+4=0
(c) 2x²+x-3=0

me ajudem pfvr é urgente ❤️​

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$-2x^{2} + 3x +5 = 0\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a} \\\frac{-3+-\sqrt{3^{2}-4*-2*(5) } }{2*(-2)}\\\frac{-3+-\sqrt{9+40} }{-4}\\\frac{-3+-\sqrt{49 } }{-4}\\\frac{-3+-7 }{-4}\\x1 = (-3+7) / 4 = 1\\x2 = (-3-7) / 4 = -\frac{10}{4} = -\frac{5}{2}$

b)

$x^{2} - 5x + 4 = 0\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a} \\\frac{-(-5)+-\sqrt{(-5)^{2}-4*1*4 } }{2*1}\\\frac{5+-\sqrt{25-16} }{2}\\\frac{5+-\sqrt{9 } }{2}\\\frac{5+-3 }{2}\\x1 = (5+3) / 2 = 4\\x2 = (5-3) / 2 = 1$

c)

$2x^{2} + x - 3 = 0\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a} \\\frac{-1+-\sqrt{1^{2}-4*2*(-3) } }{2*2}\\\frac{-1+-\sqrt{1+24} }{4}\\\frac{-1+-\sqrt{25 } }{4}\\\frac{-1+-5 }{4}\\x1 = (-1+5) / 4 = 1\\x2 = (-1-5) / 4 = -\frac{6}{4} =- \frac{3}{2}$