Question

Determine as raízes (zeros) reais de cada uma  das seguintes funções:
a) f(x) = -x² + 3/2x + 1
b) f(x) = -3x² + 6
c) f(x) =x - 2x²

Answer 1

Olá Sara,

dadas as funções do 2º grau, fazemos f(x)=0, e lembrando que:

$\begin{array}{ccc}\Delta=b^2-4ac~~~~e~~~~x= \dfrac{-b\pm \sqrt{\Delta} }{2a}\end{array}$

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$- x^{2} + \dfrac{3}{2}x+1=0\\\\ \Delta= \dfrac{3}{2}^2-4*(-1)*1\\\\\\ \Delta= \dfrac{9}{4}+4~\to~\Delta= \dfrac{25}{4}\\\\\\ x= \dfrac{- \dfrac{3}{2}\pm \sqrt{ \dfrac{25}{4} } }{2*(-1)}= \dfrac{ -\dfrac{3}{2}\pm \dfrac{5}{2} }{-2}\begin{cases}x'= \dfrac{- \dfrac{3}{2}+ \dfrac{5}{2} }{-2}= \dfrac{ \dfrac{2}{2} }{-2}= \dfrac{~~1}{-2}=-\dfrac{1}{2}\\\\ x''= \dfrac{ -\dfrac{3}{2}- \dfrac{5}{2} }{-2}= \dfrac{ -\dfrac{8}{2} }{-2}= \dfrac{-4}{-2}=2 \end{cases}\\\\\\ S=\left\{-\dfrac{1}{2},~2\right\}$

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$-3 x^{2} +6=0\\ -3 x^{2} =-6\\\\ x^{2} = \dfrac{-6}{-3}\\\\ x^{2} =2\\ x=\pm \sqrt{2}\\\\ S=\{ \sqrt{2},- \sqrt{2}\}$

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$-2 x^{2} +x=0\\ -x(2x-1)=0\\\\ -x=0~\to~x'=0\\ 2x-1=0~\to~2x=1~\to~x''= \dfrac{1}{2}\\\\ S=\left\{0,~\dfrac{1}{2}\right\}$

Espero ter ajudado e tenha ótimos estudos =))