Question

RESOLVA A EQUAÇÃO DO SEGUNDO GRAU x² - 10x + 24=0

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-10x+24=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-10~e~c=24\\\\\Delta=(b)^{2}-4(a)(c)=(-10)^{2}-4(1)(24)=100-(96)=4\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-10)-\sqrt{4}}{2(1)}=\frac{10-2}{2}=\frac{8}{2}=4\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-10)+\sqrt{4}}{2(1)}=\frac{10+2}{2}=\frac{12}{2}=6\\\\S=\{4,~6\}$

Answer 2

Equação do Segundo Grau

• Coeficientes:

$\boxed{\begin{array}{lr}\large \sf a=1\\\large \sf b=-10\\\large \sf c=24 \: \end{array}}$

• Cálculo Discriminante:

$\boxed{\begin{array}{lr}\\\large \sf \Delta={b}^{2}-4ac\\\\\large \sf \Delta = (-10)^{2} -4\cdot 1\cdot24\\\\\large \sf \Delta=100-96\\\\\large \sf \Delta=4\\\: \end{array}}$

• Bhaskara:

$\boxed{\begin{array}{lr}\\\large \sf x=\dfrac{-b \pm \sqrt{\Delta} }{2.a} \\\\ \\\large \sf x=\dfrac{-(-10) \pm \sqrt{4} }{2.1}\\\\\\\large \sf x=\dfrac{10 \pm 2 }{2}\\\: \end{array}}$

• Raízes:

$\large \boxed{\boxed{\sf x_{1}=\dfrac{10+2}{2} =\boxed{\sf 6}}}\\\\\\\large \boxed{\boxed{\sf x_{2}=\dfrac{10-2}{2} =\boxed{\sf 4}}}$

↔Resposta:

$\bullet \: \: \huge\boxed{\boxed{\huge \sf S = \{4,6\}}}$

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