Vamos resolver pela fórmula de Bhaskara:
∆ = b² – 4ac e x = (–b ± √∆)/2a
Onde "a", "b" e "c" são os coeficientes da equação
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Obs.:
- quando ∆ < 0, a equação não admite raízes reais
- quando ∆ = 0, a equação admite duas raízes reais e iguais => somente um valor para x
- quando ∆ > 0, a equação admite duas raízes reais e diferentes
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Letra A)
[tex]\begin{array}{l}\sf -x^2+3x+10=0\\\\\sf\Delta=b^2-4ac\\\\\sf \Delta=3^2-4\cdot(-1)\cdot10\\\\\sf \Delta=9+40\\\\\sf \Delta=49~~\to~~x\in\mathbb{R}~/~x'\,\neq\,x''\\\\\\\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(3)\pm\sqrt{49}}{2\cdot(-1)}~~\Rightarrow~~x=\dfrac{-3\pm7}{-2~}\\\\\Rightarrow~\begin{cases}\sf x'=\dfrac{-3+7}{-2~}=\dfrac{4}{-2~}=-2\\\\\sf x''=\dfrac{-3-7}{-2~}=\dfrac{-10~}{-2~}=5\end{cases}\end{array}[/tex]
Assim o conjunto solução é:
[tex]\large\begin{array}{l}\boldsymbol{\sf S=\Big\{-2~~;~~5\Big\}}\end{array}[/tex]
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Letra B)
[tex]\begin{array}{l}\sf x^2-10x+24=0\\\\\sf\Delta=b^2-4ac\\\\\sf \Delta=(-10)^2-4\cdot1\cdot24\\\\\sf \Delta=100-96\\\\\sf \Delta=4~~\to~~x\in\mathbb{R}~/~x'\,\neq\,x'' \\ \\ \\ \sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-10)\pm\sqrt{4}}{2\cdot1}~~\Rightarrow~~x=\dfrac{10\pm2}{2}\\\\\Rightarrow~\begin{cases}\sf x'=\dfrac{10+2}{2}=\dfrac{12}{2}=6\\\\\sf x''=\dfrac{10-2}{2}=\dfrac{8}{2}=4\end{cases}\end{array}[/tex]
Assim o conjunto solução é:
[tex]\large\begin{array}{l}\boldsymbol{\sf S=\Big\{4~~;~~6\Big\}}\end{array}[/tex]
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Letra C)
[tex]\begin{array}{l}\sf 3x^2-8x+10=0\\\\\sf\Delta=b^2-4ac\\\\\sf \Delta=(-8)^2-4\cdot3\cdot10\\\\\sf \Delta=64-120\\\\\sf \Delta=-56~~\to~~x\notin\mathbb{R}\end{array}[/tex]
Assim o conjunto solução é vazio:
[tex]\large\begin{array}{l}\boldsymbol{\sf S=\Big\{\,\,\,\Big\}}\end{array}[/tex]
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Letra D)
[tex]\begin{array}{l}\sf x^2-4x+4=0\\\\\sf\Delta=b^2-4ac\\\\\sf \Delta=(-4)^2-4\cdot1\cdot4\\\\\sf \Delta=16-16\\\\\sf \Delta=0~~\to~~x\in\mathbb{R}~/~x'=x''\\\\\\\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-4)\pm\sqrt{0}}{2\cdot1}~~\Rightarrow~~x=\dfrac{4\pm0}{2}\\\\\Rightarrow~\begin{cases}\sf x'=x''=\dfrac{4}{2}=2\end{cases}\end{array}[/tex]
Assim o conjunto solução é:
[tex]\large\begin{array}{l}\boldsymbol{\sf S=\Big\{\,2\,\Big\}}\end{array}[/tex]
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Att. Nasgovaskov
