Question

Determine os zeros das funções quadráticas definidas por:
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Answer 1

Explicação passo-a-passo:

a)

$\sf \dfrac{1}{2}x^2-4x+6=0$

$\sf \Delta=(-4)^2-4\cdot\dfrac{1}{2}\cdot6$

$\sf \Delta=16-12$

$\sf \Delta=4$

$\sf x=\dfrac{-(-4)\pm\sqrt{4}}{2\cdot\frac{1}{2}}=\dfrac{4\pm2}{1}$

$\sf x'=\dfrac{4+2}{1}~\rightarrow~x'=\dfrac{6}{1}~\rightarrow~x'=6$

$\sf x"=\dfrac{4-2}{1}~\rightarrow~x"=\dfrac{2}{1}~\rightarrow~x"=2$

Os zeros dessa função são $\sf 6$ e $\sf 2$

b)

$\sf 2x^2-2x-40=0$

$\sf \Delta=(-2)^2-4\cdot2\cdot(-40)$

$\sf \Delta=4+320$

$\sf \Delta=324$

$\sf x=\dfrac{-(-2)\pm\sqrt{324}}{2\cdot2}=\dfrac{2\pm18}{4}$

$\sf x'=\dfrac{2+18}{4}~\rightarrow~x'=\dfrac{20}{4}~\rightarrow~x'=5$

$\sf x"=\dfrac{2-18}{4}~\rightarrow~x"=\dfrac{-16}{4}~\rightarrow~x"=-4$

Os zeros dessa função são $\sf 5$ e $\sf -4$