Question

Resolva as equações de 2o grau:
1) X2 - 5x + 6 = 0

2) X2 - 9x + 20 = 0

3) X2 + 9x + 14 = 0

4) X2 - 3x – 10 = 0

Answer 1

Explicação passo-a-passo:

1) $\sf x^2-5x+6=0$

$\sf \Delta=(-5)^2-4\cdot1\cdot6$

$\sf \Delta=25-24$

$\sf \Delta=1$

$\sf x=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}=\dfrac{5\pm1}{2}$

$\sf x'=\dfrac{5+1}{2}~\Rightarrow~x'=\dfrac{6}{2}~\Rightarrow~\red{x'=3}$

$\sf x"=\dfrac{5-1}{2}~\Rightarrow~x"=\dfrac{4}{2}~\Rightarrow~\red{x"=2}$

O conjunto solução é $\sf S=\{2,3\}$

2) $\sf x^2-9x+20=0$

$\sf \Delta=(-9)^2-4\cdot1\cdot20$

$\sf \Delta=81-80$

$\sf \Delta=1$

$\sf x=\dfrac{-(-9)\pm\sqrt{1}}{2\cdot1}=\dfrac{9\pm1}{2}$

$\sf x'=\dfrac{9+1}{2}~\Rightarrow~x'=\dfrac{10}{2}~\Rightarrow~\red{x'=5}$

$\sf x"=\dfrac{9-1}{2}~\Rightarrow~x"=\dfrac{8}{2}~\Rightarrow~\red{x"=4}$

O conjunto solução é $\sf S=\{4,5\}$

3) $\sf x^2-9x+14=0$

$\sf \Delta=(-9)^2-4\cdot1\cdot14$

$\sf \Delta=81-56$

$\sf \Delta=25$

$\sf x=\dfrac{-(-9)\pm\sqrt{25}}{2\cdot1}=\dfrac{9\pm5}{2}$

$\sf x'=\dfrac{9+5}{2}~\Rightarrow~x'=\dfrac{14}{2}~\Rightarrow~\red{x'=7}$

$\sf x"=\dfrac{9-5}{2}~\Rightarrow~x"=\dfrac{4}{2}~\Rightarrow~\red{x"=2}$

O conjunto solução é $\sf S=\{2,7\}$

4) $\sf x^2-3x-10=0$

$\sf \Delta=(-3)^2-4\cdot1\cdot(-10)$

$\sf \Delta=9+40$

$\sf \Delta=49$

$\sf x=\dfrac{-(-3)\pm\sqrt{49}}{2\cdot1}=\dfrac{3\pm7}{2}$

$\sf x'=\dfrac{3+7}{2}~\Rightarrow~x'=\dfrac{10}{2}~\Rightarrow~\red{x'=5}$

$\sf x"=\dfrac{3-7}{2}~\Rightarrow~x"=\dfrac{-4}{2}~\Rightarrow~\red{x"=-2}$

O conjunto solução é $\sf S=\{-2,5\}$