Question

equaçao do 2 grau
a;x²-5x +6=0
b;x²-8x+12=0
c;x²-2x-8=0
d;x²-5x+8=0

Answer 1

a) $x^2-5x+6=0~~~\Rightarrow~~\left\{\! \begin{array}{l}a=1\\b=-5\\c=6 \end{array} \right.$

$\Delta=b^2-4ac\\\\ \Delta=(-5)^2-4\cdot 1\cdot 6\\\\ \Delta=25-24\\\\ \Delta=1\\\\\\ x=\dfrac{-b \pm \sqrt{\Delta}}{2a}\\\\\\ x=\dfrac{-(-5) \pm \sqrt{1}}{2\cdot 1}$

$x=\dfrac{5 \pm 1}{2}\\\\\\ \begin{array}{rcl} x=\dfrac{5-1}{2}&~\text{ ou }~&x=\dfrac{5+1}{2}\\\\\\ x=\dfrac{4}{2}&~\text{ ou }~&x=\dfrac{6}{2} \end{array}\\\\\\ ~~~~~~\boxed{ \begin{array}{rcl} x=2&~\text{ ou }~&x=3 \end{array} }$

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b) $x^2-8x+12=0~~~\Rightarrow~~\left\{\! \begin{array}{l}a=1\\b=-8\\c=12 \end{array} \right.$

$\Delta=b^2-4ac\\\\ \Delta=(-8)^2-4\cdot 1\cdot 12\\\\ \Delta=64-48\\\\ \Delta=16\\\\\\ x=\dfrac{-b \pm \sqrt{\Delta}}{2a}\\\\\\ x=\dfrac{-(-8) \pm \sqrt{16}}{2\cdot 1}$

$x=\dfrac{8 \pm 4}{2}\\\\\\ x=\dfrac{\diagup\!\!\!\! 2\cdot (4 \pm 2)}{\diagup\!\!\!\! 2}\\\\\\ x=4 \pm 2\\\\ \begin{array}{rcl} x=4-2&~\text{ ou }~&x=4+2 \end{array}\\\\ ~~~~~\,\boxed{ \begin{array}{rcl} x=2&~\text{ ou }~&x=6 \end{array} }$

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c) $x^2-2x-8=0~~~\Rightarrow~~\left\{\! \begin{array}{l}a=1\\b=-2\\c=-8 \end{array} \right.$

$\Delta=b^2-4ac\\\\ \Delta=(-2)^2-4\cdot 1\cdot (-8)\\\\ \Delta=4+32\\\\ \Delta=36\\\\\\ x=\dfrac{-b \pm \sqrt{\Delta}}{2a}\\\\\\ x=\dfrac{-(-2) \pm \sqrt{36}}{2\cdot 1}$

$x=\dfrac{2 \pm 6}{2}\\\\\\ x=\dfrac{\diagup\!\!\!\! 2\cdot (1\pm 3)}{\diagup\!\!\!\! 2}\\\\\\ x=1\pm 3\\\\ \begin{array}{rcl} x=1-3&~\text{ ou }~&x=1+3 \end{array}\\\\ ~~~\boxed{ \begin{array}{rcl} x=-2&~\text{ ou }~&x=4 \end{array} }$

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d) $x^2-5x+8=0~~~\Rightarrow~~\left\{\! \begin{array}{l}a=1\\b=-5\\c=8 \end{array} \right.$

$\Delta=b^2-4ac\\\\ \Delta=(-5)^2-4\cdot 1\cdot 8\\\\ \Delta=25-32\\\\ \Delta=-7<0$


Como o discriminante $\Delta$ é negativo, a equação dada não possui solução real.

O conjunto solução é vazio: $S=\varnothing.$


Bons estudos! :-)