Question

O produto das raízes reais da equação |x²-3x+2| = |2x-3|

Answer 1

Resolvendo a equação modular:

$|x^2-3x+2|=|2x-3|\\\\ x^2-3x+2=\pm (2x-3)\\\\ \begin{array}{rcl} x^2-3x+2=-(2x-3)&~\text{ ou }~&x^2-3x+2=2x-3\\\\ x^2-3x+2=-2x+3&~\text{ ou }~&x^2-3x+2=2x-3\\\\ x^2-3x+2+2x-3=0&~\text{ ou }~&x^2-3x+2-2x+3=0\\\\ x^2-x-1=0&~\text{ ou }~&x^2-5x+5=0 \end{array}$


O problema consiste em resolver as duas equações do 2º grau acima:

$\bullet\;\;x^2-x-1=0~~\Rightarrow~~\left\{\! \begin{array}{l}a=1\\b=-1\\c=-1 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=(-1)^2-4\cdot 1\cdot (-1)\\\\ \Delta=1+4\\\\ \Delta=5\\\\\\ x=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ x=\dfrac{-(-1)\pm \sqrt{5}}{2\cdot 1}$

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


Então, as duas primeiras soluções para a equação modular inicial são

$\boxed{\begin{array}{rcl} x_1=\dfrac{1-\sqrt{5}}{2}&~\text{ e }~&x_2=\dfrac{1+\sqrt{5}}{2} \end{array}}$

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$\bullet\;\;x^2-5x+5=0~~\Rightarrow~~\left\{\! \begin{array}{l}a=1\\b=-5\\c=5 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=(-5)^2-4\cdot 1\cdot 5\\\\ \Delta=25-20\\\\ \Delta=5\\\\\\ x=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ x=\dfrac{-(-5)\pm \sqrt{5}}{2\cdot 1}$

$x=\dfrac{5\pm \sqrt{5}}{2}\\\\\\ \begin{array}{rcl} x=\dfrac{5-\sqrt{5}}{2}&~\text{ ou }~&x=\dfrac{5+\sqrt{5}}{2} \end{array}$


As duas outras soluções para a equação modular dada inicialmente são

$\boxed{\begin{array}{rcl} x_3=\dfrac{5-\sqrt{5}}{2}&~\text{ e }~&x_4=\dfrac{5+\sqrt{5}}{2} \end{array}}$

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Queremos o produto das soluções, então queremos calcular

$x_1\cdot x_2\cdot x_3\cdot x_4\\\\ =\dfrac{1-\sqrt{5}}{2}\cdot \dfrac{1+\sqrt{5}}{2}\cdot \dfrac{5-\sqrt{5}}{2}\cdot \dfrac{5+\sqrt{5}}{2}\\\\\\ =\dfrac{\big(1-\sqrt{5}\big)\cdot \big(1+\sqrt{5}\big)}{2\cdot 2}\cdot \dfrac{\big(5-\sqrt{5}\big)\cdot \big(5+\sqrt{5}\big)}{2\cdot 2}\\\\\\ =\dfrac{1^2-\big(\sqrt{5}\big)^2}{4}\cdot \dfrac{5^2-\big(\sqrt{5}\big)^2}{4}\\\\\\ =\dfrac{1-5}{4}\cdot \dfrac{25-5}{4}\\\\\\ =\dfrac{-4}{4}\cdot \dfrac{20}{4}\\\\\\ =-1\cdot 5\\\\ =\boxed{\begin{array}{c}-5 \end{array}}$


Bons estudos! :-)