Question

determine os zeros das funcoes definidas por f(x)=x^2-5/3x-2/3f(x)=4x^2-3x-1f(x)=-x^2 4? me ajudeeem por favor!

Answer 1

a) $f(x)=x^2-\dfrac{5x}{3}-\dfrac{2}{3}$

$x^2-\dfrac{5x}{3}-\dfrac{2}{3}=0$

$3x^2-5x-2=0$

$\Delta=(-5)^2-4\cdot3\cdot(-2)=25+24=49$

$x=\dfrac{-(-5)\pm\sqrt{49}}{2\cdot3}=\dfrac{5\pm7}{6}$

$x'=\dfrac{5+7}{6}=\dfrac{12}{6}=2$

$x"=\dfrac{5-7}{6}=\dfrac{-2}{6}=-\dfrac{1}{3}$


b) $f(x)=4x^2-3x-1$

$4x^2-3x-1=0$

$\Delta=(-3)^2-4\cdot4\cdot(-1)=9+16=25$

$x=\dfrac{-(-3)\pm\sqrt{25}}{2\cdot4}=\dfrac{3\pm5}{8}$

$x'=\dfrac{3+5}{8}=\dfrac{8}{8}=1$

$x"=\dfrac{3-5}{8}=\dfrac{-2}{8}=-\dfrac{1}{4}$


c) $f(x)=-x^2+4$

$-x^2+4=0$

$x^2=4$

$x=\pm\sqrt{4}$

$x'=-2$

$x"=2$