Question

Pessoal se puderem me ajudar eu agradeço 

91f1031acbea1a1dfc038635b11273d1.docx

Answer 1

1) $x^2+2x+1=2x^2-2x+1$

$2x^2-x^2-2x-2x+1-1=0$

$x^2-4x=0$

$\Delta=(-4)^2-4\cdot1\cdot0=16$

$x=\dfrac{-(-4)\pm\sqrt{16}}{2}=\dfrac{4\pm4}{2}=2\pm2$

$x'=2+2=4$ e $x"=2-2=0$

As raízes são 4 e 0.


2) $x(x+11)+2(x+21)=0$

$x^2+11x+2x+42=0$

$x^2+13x+42=0$

$\Delta=13^2-4\cdot1\cdot42=169=168=1$

$x=\dfrac{-13\pm\sqrt{1}}{2}=\dfrac{-13\pm1}{2}$

$x'=\dfrac{-13+1}{2}=\dfrac{-12}{2}=-6$

$x"=\dfrac{-13-1}{2}=\dfrac{-14}{2}=-7$

$S=\{-6,-7\}$.

5) $x^2+(x+1)^2=(x+2)^2$

$x^2+x^2+2x+1=x^2+4x+4$

$2x^2+2x+1=x^2+4x+4$

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

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

$\Delta=(-2)^2-4\cdot1\cdot(-3)=4+12=16$

$x=\dfrac{-(-2)\pm\sqrt{16}}{2}=\dfrac{2\pm4}{2}=1\pm2$

$x'=1+2=3$ e $x"=1-2=-1$ (não satisfaz)

$x=3$, $x+1=4$ e $x+2=5$.


6) $\begin{cases} x+y=9 \\ xy=14\end{cases}$

$x=9-y$

$(9-y)y=14$

$9y-y^2-14=0$

$y^2-9y+14=0$

$\Delta=(-9)^2-4\cdot1\cdot14=81-56=25$

$y=\dfrac{-(-9)\pm\sqrt{25}}{2}=\dfrac{9\pm5}{2}$

$y='\dfrac{9+5}{2}=\dfrac{14}{2}=7$

$x'=9-7=2$

$y"=\dfrac{9-5}{2}=\dfrac{4}{2}=2$

$x"=9-2=7$

$(x,y)=(7,2)$ ou $(x,y)=(2,7)$.

7) $4x^2-3px+p-4=0$

Soma das raízes:

$S=\dfrac{-b}{a}=\dfrac{3p}{4}$

Produto das raízes:

$P=\dfrac{c}{a}=\dfrac{p-4}{4}$

Igualando:

$\dfrac{3p}{4}=\dfrac{p-4}{4}$

$3p=p-4$

$2p=-4$

$p=-2$

8) Igual a anterior.

9) $x^2-11x+28=0$

$S=\dfrac{-b}{a}=\dfrac{-(-11)}{1}=11$

$P=\dfrac{c}{a}=\dfrac{28}{1}=28$

$S-P=11-28=-17$


10) $x^2-0,8x-1,6=0$

$S=\dfrac{-b}{a}=\dfrac{-(-0,8)}{1}=0,8$

$P=\dfrac{c}{a}=\dfrac{-1,6}{1}=-1,6$

$\dfrac{S}{P}=\dfrac{0,8}{-1,6}=-0,5$.