Question

determine as raízes ( zeros ) reais de cada uma das funções seguintes . ​

Answer 1

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$\tt{a)}~\sf{y=2x^2-3x+1}\\\sf{a=2~~b=-3~~c=1}\\\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\\\sf{x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4\cdot2\cdot1}}{2\cdot1}}\\\sf{x=\dfrac{3\pm\sqrt{9-8}}{2}}\\\sf{x=\dfrac{3\pm\sqrt{1}}{2}}\\\sf{x=\dfrac{3\pm1}{2}}\begin{cases}\sf{x_1=\dfrac{3+1}{2}=\dfrac{4}{2}=2}\\\sf{x_2=\dfrac{3-1}{2}=\dfrac{2}{2}=1}\end{cases}$

$\tt{b)}~\sf{y=4x-x^2}\\\sf{a=-1~~b=4~~c=0}\\\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\\\sf{x=\dfrac{-(4)\pm\sqrt{(4)^2-4\cdot(-1)\cdot0}}{2\cdot(-1)}}\\\sf{x=\dfrac{-4\pm\sqrt{16}}{-2}}\\\sf{x=\dfrac{-4\pm4}{-2}}\begin{cases}\sf{x_1=\dfrac{-4+4}{2}=\dfrac{0}{-2}=0}\\\sf{x_2=\dfrac{-4-4}{-2}=\dfrac{-8}{-2}=4}\end{cases}$

$\tt{c)}~\sf{y=-x^2+2x+15}\\\sf{a=-1~~b=2~~c=15}\\\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\\\sf{x=\dfrac{-(2)\pm\sqrt{(2)^2-4\cdot(-1)\cdot15}}{2\cdot(-1)}}\\\sf{x=\dfrac{-2\pm\sqrt{4+60}}{-2}}\\\sf{x=\dfrac{2\pm\sqrt{64}}{-2}}\\\sf{x=\dfrac{2\pm8}{-2}}\begin{cases}\sf{x_1=\dfrac{2+8}{-2}=\dfrac{10}{-2}=-5}\\\sf{x_2=\dfrac{2-8}{-2}=\dfrac{-6}{-2}=3}\end{cases}$

$\tt{d)}~\sf{y=9x^2-1}\\\sf{a=9~~b=0~~c=-1}\\\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\\\sf{x=\dfrac{0\pm\sqrt{0^2-4\cdot9\cdot(-1)}}{2\cdot9}}\\\sf{x=\dfrac{0\pm\sqrt{0+36}}{18}}\\\sf{x=\dfrac{0\pm\sqrt{36}}{18}}\\\sf{x=\dfrac{0\pm6}{18}}\begin{cases}\sf{x_1=\dfrac{0+6}{18}=\dfrac{6\div6}{18\div6}=\dfrac{1}{3}}\\\sf{x_2=\dfrac{0-6}{18}=\dfrac{-6\div6}{18\div6}=-\dfrac{1}{3}}\end{cases}$

$\tt{e)}~\sf{y=-x^2+6x-9}\\\sf{a=-1~~b=6~~c=-9}\\\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\\\sf{x=\dfrac{-(6)\pm\sqrt{(6)^2-4\cdot(-1)\cdot(-9)}}{2\cdot(-1)}}\\\sf{x=\dfrac{-6\pm\sqrt{36-36}}{-2}}\\\sf{x=\dfrac{-6\pm\sqrt{0}}{-2}}\\\sf{x=\dfrac{-6\pm0}{-2}}\begin{cases}\sf{x_1=\dfrac{-6+0}{-2}=\dfrac{-6}{-2}=3}\\\sf{x_2=\dfrac{-6-0}{-2}=\dfrac{-6}{-2}=3}\end{cases}$