Question

determine as raizes das funções:
a) y = x² - 7x + 12
b) y = 2x² + 5x - 3
c) y = - x² - x + 30​

Answer 1

Resposta:

a) x^2 - 7x + 12 = 0 <=> x=$\frac{7+/- \sqrt{7^{2}-4*1*12 } }{2*1}$ <=> x= $\frac{7+/-\sqrt{49-48} }{2}$ <=> x= $\frac{7+/- \sqrt{1} }{2}$ <=> x= $\frac{7+1}{2}$ V x=$\frac{7-1}{2}$ <=> x= $\frac{8}{2}$ V x= $\frac{6}{2}$ <=> x= 4 V x=3

b) 2x^2 + 5x - 3 =0 <=> x= $\frac{-5+/- \sqrt{5^{2}-4*2*(-3) } }{2*2}$ <=> x=$\frac{-5+/-\sqrt{25+24} }{4}$ <=>             x= $\frac{-5+/-\sqrt{49} }{4}$ <=> x= $\frac{-5+7}{4}$ V x= $\frac{-5-7}{4}$ <=> x= $\ \frac{2}{4}$ V x= $\frac{-12}{4}$ <=> x=$\frac{1}{2}$ V x= -3

c) -x^2 -x + 30 =0 <=> $\frac{1+/-\sqrt{(-1)^{2} -4*(-1)*(30)} }{2*(-1)}$ <=> x= $\frac{1+/- \sqrt{1+120} }{-2}$ <=> x= $\frac{1+11}{-2}$ V x= $\frac{1-11}{-2}$ <=> x= $\frac{12}{-2}$ V x=$\frac{-10}{-2}$ <=> x= -6 V x = 5

Explicação passo a passo:

raízes das funções ou zeros das funções

Nos casos ax^2+bx+c casos utiliza-se a formula resolvente, sendo essa

$\frac{-b+/-\sqrt{b^{2} -4*a*c } }{2*a}$