Question

Determine, se existirem as raizes de cada função

a) F (x) = 3 x²- 7x + 2

b) F (x) = x²+3x-4

c) F (x) = x² + 6x +5

d) F (x) = x²- 6x+9

Por favor preciso desta lição URGENTE para amanhã se não estou REPROVADO....

Answer 1

a) $f(x)=3x^2-7x+2$

$3x^2-7x+2=0$

$\Delta=(-7)^2-4\cdot3\cdot2=49-24=25$

$x=\dfrac{-(-7)\pm\sqrt{25}}{2\cdot3}=\dfrac{7\pm5}{6}$

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

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

As raízes são $\dfrac{1}{3}$ e $2$

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

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

$\Delta=3^2-4\cdot1\cdot(-4)=9+16=25$

$x=\dfrac{-3\pm\sqrt{25}}{2\cdot1}=\dfrac{-3\pm5}{2}$

$x'=\dfrac{-3+5}{2}=\dfrac{2}{2}=1$

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

As raízes são $-4$ e $1$

c) $f(x)=x^2+6x+5$

$x^2+6x+5=0$

$\Delta=6^2-4\cdot1\cdot5=36-20=16$

$x=\dfrac{-6\pm\sqrt{16}}{2\cdot1}=\dfrac{-6\pm4}{2}=-3\pm2$

$x'=-3+2=-1$

$x"=-3-2=-5$

As raízes são $-5$ e $-1$

d) $f(x)=x^2-6x+9$

$x^2-6x+9=0$

$\Delta=(-6)^2-4\cdot1\cdot9=36-36=0$

$x=\dfrac{-(-6)\pm\sqrt{0}}{2\cdot1}=\dfrac{6}{2}=3$

A raiz única dessa função é $3$