Question

50 pontos me ajudem
Determine o conjunto solução de cada uma das seguintes equações do 2°grau:

a) (x+2)²+x=0
b) 3x=2(x-1)+3
c) x(x+11)+2(x+21)=0
d) 6(x²-1)-14=5x²+x

Answer 1

$a) \\ (x+2)^2+x=0 \\ \\ (x + 2)(x+ 2) + x \\ \\ x^2 + 2x + 2x + 4 + x = 0 \\ \\ x^2 + 5x + 4 = 0$

Resolvendo por Bháskara:

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=5,c=4
Δ= b^2 - 4ac
Δ= 5^2 - 4*1*4
Δ= 25 - 16
Δ= 9

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-5 \pm \sqrt{9}}{2*1} \\ \\ \\ x = \dfrac{-5 \pm 3}{2} \\ \\ \\ x' = \dfrac{-5 + 3}{2} \\ \\ \\ x' = \dfrac{-2}{2} \\ \\ \\ x' = -1 \\ \\ \\ x'' = \dfrac{-5 - 3}{2} \\ \\ \\ x'' = \dfrac{-8}{2} \\ \\ \\ x'' = -4$

S = {-1, -4}

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$b) \\ 3x=2(x-1)+3 \\ \\ 3x = 2x - 2 + 3 \\ \\ 3x - 2x - 2 + 3 = 0$

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$c) \\ x(x+11)+2(x+21)=0 \\ \\ x^2 + 11x + 2x + 42 \\ \\ x^2 + 13x + 42 = 0$

a=1, b=13,c=42
Δ= b^2 - 4ac
Δ= 13^2 - 4*1*42
Δ= 169 - 168
Δ= 1

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-13 \pm \sqrt{1}}{2*1} \\ \\ \\ x = \dfrac{-13 \pm 1}{2} \\ \\ \\ x' = \dfrac{-13 + 1}{2} \\ \\ \\ x' = \dfrac{-12}{2} \\ \\ \\ \\ x' = -6 \\ \\ \\ x'' = \dfrac{-13 - 1}{2} \\ \\ \\ x'' = \dfrac{-14}{2} \\ \\ \\ x'' = -7$

S = {-6, -7}
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$d) \\ \\ 6(x^2-1)-14=5x^2+x \\ \\ 6x^2 - 6 - 14 = 5x^2 + x \\ \\ 6x^2 -20 = 5x^2 + x \\ \\ x^2 -x - 20 = 0$
$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

a=1, b=-1,c=-20
Δ= b^2 - 4ac
Δ= (-1)^2 - 4*1*-20
Δ= 1 + 80
Δ= 81

$x = \dfrac{-(-1) \pm \sqrt{81}}{2*1} \\ \\ \\ x = \dfrac{1 \pm 9}{2} \\ \\ \\ x' = \dfrac{1 + 9}{2} \\ \\ \\ x' = \dfrac{10}{2} \\ \\ \\ x' = 5 \\ \\ \\ x'' = \dfrac{1 - 9}{2} \\ \\ \\ x'' = \dfrac{-8}{2} \\ \\ \\ x'' = -4$

S = {5, -4}

Answer 2

a) $(x+2)^2+x=0$

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

$x^2+5x+4=0$

$\Delta=5^2-4\cdot1\cdot4=25-16=9$

$x=\dfrac{-5\pm\sqrt{9}}{2}=\dfrac{-5\pm3}{2}$

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

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

$S=\{-4,-1\}$

b) $3x=2(x-1)^2+3$

$3x=2(x^2-2x+1)+3$

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

$2x^2-7x+5=0$

$\Delta=(-7)^2-4\cdot2\cdot5=49-40=9$

$x=\dfrac{-(-7)\pm\sqrt{9}}{2\cdot2}=\dfrac{7\pm3}{4}$

$x'=\dfrac{7+3}{4}=\dfrac{10}{4}=\dfrac{5}{2}$

$x"=\dfrac{7-3}{4}=1$

$S=\{1,\frac{5}{2}\}$.


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

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

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

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

$x=\dfrac{-13\pm1}{2}$

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

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

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


d) $6(x^2-1)-14=5x^2+x$

$6x^2-6-14-5x^2-x=0$

$x^2-x-20=0$

$\Delta=(-1)^2-4\cdot1\cdot(-20)=1+80=81$

$x=\dfrac{-(-1)\pm\sqrt{81}}{2}=\dfrac{1\pm9}{2}$

$x'=\dfrac{1+9}{2}=5$

$x"=\dfrac{1-9}{2}=-4$

$S=\{-4,5\}$