Question

3) Resolva as equações do 2o grau completas usando a fórmula de Bhaskara:
a) 6x²+x-1=0 b) 3x²-7x+2=0
c) x²-4x-5=0 d) x²-5x+6=0

Answer 1

Item a)

Em $6x^2 +x-5 = 0$, temos  $a = 6,\ b = 1\mbox{ e } c = -1.$ Então, o discriminante será dado por

$\begin{array}{rcl}\Delta &=& b^2-4ac \\ \\ \Delta &=& 1^2 - 4 \cdot 6 \cdot (-1) \\ \\ \Delta &=& 1 +24 \\ \\ \Delta &=& 25\end{array}$

Portanto,

$\begin{array}{rcl}x &=&\dfrac{-b\pm\sqrt{\Delta}}{2a} \\ \\ x &=& \dfrac{-1\pm \sqrt{25}}{2\cdot 6}\\ \\ x &=& \dfrac{-1\pm5}{12} \end{array}$

Assim, temos

$x_1 = \dfrac{-1+5}{12} = \dfrac{4}{12} = \dfrac{1}{3}.$

$x_2 = \dfrac{-1-5}{12} = \dfrac{-6}{12} = -\dfrac{1}{2}.$.

Item b)

Em $3x^2 -7x+2= 0$, temos  $a = 3,\ b = -7\mbox{ e } c = 2.$ Então, o discriminante será dado por

$\begin{array}{rcl}\Delta &=& b^2-4ac \\ \\ \Delta &=& (-7)^2 - 4 \cdot 3 \cdot 2 \\ \\ \Delta &=& 49-24 \\ \\ \Delta &=& 25\end{array}$

Portanto,

$\begin{array}{rcl}x &=&\dfrac{-b\pm\sqrt{\Delta}}{2a} \\ \\ x &=& \dfrac{-(-7)\pm \sqrt{25}}{2\cdot 3}\\ \\ x &=& \dfrac{7\pm5}{6} \end{array}$

Assim, temos

$x_1 = \dfrac{7+5}{6} = \dfrac{12}{6} = 2.$

$x_2 = \dfrac{7-5}{6} = \dfrac{2}{6} = \dfrac{1}{3}.$.

Item c)

Em $x^2 -4x+5= 0$, temos  $a = 1,\ b = -4\mbox{ e } c = 5.$ Então, o discriminante será dado por

$\begin{array}{rcl}\Delta &=& b^2-4ac \\ \\ \Delta &=& (-4)^2 - 4 \cdot 1 \cdot (-5) \\ \\ \Delta &=& 16+20\\ \\ \Delta &=& 36\end{array}$

Portanto,

$\begin{array}{rcl}x &=&\dfrac{-b\pm\sqrt{\Delta}}{2a} \\ \\ x &=& \dfrac{-(-4)\pm \sqrt{36}}{2\cdot 1}\\ \\ x &=& \dfrac{4\pm6}{2} \end{array}$

Assim, temos

$x_1 = \dfrac{4+6}{2} = \dfrac{10}{2} = 5.$

$x_2 = \dfrac{4-6}{2} = -\dfrac{2}{2} = -1.$.

Item c)

Em $x^2 -5x+6= 0$, temos  $a = 1,\ b = -5\mbox{ e } c = 6.$ Então, o discriminante será dado por

$\begin{array}{rcl}\Delta &=& b^2-4ac \\ \\ \Delta &=& (-5)^2 - 4 \cdot 1 \cdot 6 \\ \\ \Delta &=& 25-25\\ \\ \Delta &=& 1\end{array}$

Portanto,

$\begin{array}{rcl}x &=&\dfrac{-b\pm\sqrt{\Delta}}{2a} \\ \\ x &=& \dfrac{-(-5)\pm \sqrt{1}}{2\cdot 1}\\ \\ x &=& \dfrac{5\pm1}{2} \end{array}$

Assim, temos

$x_1 = \dfrac{5+1}{2} = \dfrac{6}{2} = 3.$

$x_2 = \dfrac{5-1}{2} = -\dfrac{4}{2} = 2.$

Ufa! Espero ter ajudado :D