Question

Sejam x₁ e x₂ as raízes da equação 2x²-√6x + P-2= 0. Se (x₁ + x₂)²= x₁ × x₂, então P é igual?

Answer 1

$2x^{2}-\sqrt{6}x+P-2=0\\\\a=2\\b=-\sqrt{6}\\c=P-2$
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$(x_{1}+x_{2})^{2}=x_{1}\cdot x_{2}\\\\S^{2}=P\\\\\\\left(-\dfrac{b}{a}\right)^{2}=\dfrac{c}{a}\\\\\\\dfrac{b^{2}}{a^{2}}=\dfrac{c}{a}\\\\\\\dfrac{b^{2}}{a}=c$

Como a = 2, b = - √6 e c = P - 2:

$\dfrac{(-\sqrt{6})^{2}}{2}=P-2\\\\\\P-2=\dfrac{6}{2}\\\\\\P-2=3\\\\\\P=3+2\\\\\\\boxed{\boxed{P=5}}$

Answer 2

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$2 x^{2} - \sqrt{6}x+p-2=0 \\ a=2 \\ b=- \sqrt{6} \\ c=P-2 \\ \\ x'+x"=- \frac{b}{a}= \frac{ \sqrt{6} }{2} \\ \\ x'.x"= \frac{c}{a} = \frac{P-2}{2}$

$(x'+x")^2=x'.x" \\ \\ ( \frac{ \sqrt{6} }{2} )^2= \frac{P-2}{2} \\ \\ \frac{6}{4} = \frac{P-2}{2} \\ \\ 4(P-2)=6.2 \\ 4P-8=12 \\ 4P=12+8 \\ 4P=20 \\ P=20\div4 \\ P=5$