Question

dado a equação de 2° grau: 2x .(x - 4) + 10 = - x² + 3x​

Answer 1

As Raízes da equação 3x² - 11x + 10 = 0 é; S = {2, 5/3}

• Resolvendo  simplificando.

$\boxed{\begin{array}{lr} 2x.(x-4)+10=-x^2+3x\\2x^2 - 8x +10= -x^2 + 3x\\2x^2+x^2-3x-8x+10=0\\\boxed{3x^2-11x+10=0} \end{array}}$

• Coeficientes.

$\boxed{\begin{array}{lr} 3x^2-11x+10=0 \rightarrow\begin{cases} a=3\\b=-11\\c=10 \end{cases} \end{array}}$

• Resolvendo.

$\boxed{\begin{array}{lr} x=\dfrac{11\pm\sqrt{\Delta}}{2.3} \end{array}}\to \boxed{\begin{array}{lr} \Delta=(-11)^2-4.3.10 \end{array}}$

• Resolvendo o Delta.

$\boxed{\begin{array}{lr} \Delta=b^2-4.a.c\\\Delta=(-11)^2-4.3.10\\\Delta=121-120\\\Delta=1 \end{array}}$

• Valor do discriminante é um.

• Agora basta resolver.

$\boxed{\begin{array}{lr} x=\dfrac{11\pm\sqrt{1}}{2.3} \end{array}}\to \boxed{\begin{array}{lr} \boxed{\begin{array}{lr} x=\dfrac{11\pm1}{6} \end{array}}\end{array}}$

• Retirando o mais ou menos.

$\boxed{\begin{array}{lr} x'=\dfrac{11+1}{6} \end{array}}>\boxed{\begin{array}{lr} x'=\dfrac{12}{6} \end{array}}>\boxed{\begin{array}{lr} \boxed{\begin{array}{lr} x'=2\ \ \checkmark \end{array}} \end{array}}\\\\\\\boxed{\begin{array}{lr} x''=\dfrac{11-1}{6} \end{array}}>\boxed{\begin{array}{lr} x''=\dfrac{10}{6} \end{array}}>\boxed{\begin{array}{lr} \boxed{\begin{array}{lr} x''=\dfrac{5}{3}\ \ \checkmark \end{array}} \end{array}}$

Resposta;

S = {2, 5/3}

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