Question

Dada a equação 2x² - 5x + 2 = 0. Descubra os zeros da equação *

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Answer 1

Explicação passo-a-passo:

$\sf 2x^2-5x+2=0$

$\sf \Delta=(-5)^2-4\cdot2\cdot2$

$\sf \Delta=25-16$

$\sf \Delta=9$

$\sf x=\dfrac{-(-5)\pm\sqrt{9}}{2\cdot2}=\dfrac{5\pm3}{4}$

$\sf x'=\dfrac{5+3}{4}~\Rightarrow~x'=\dfrac{8}{4}~\Rightarrow~\red{x'=2}$

$\sf x"=\dfrac{5-3}{4}~\Rightarrow~x"=\dfrac{2}{4}~\Rightarrow~\red{x"=\dfrac{1}{2}}$

Answer 2

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~2x^{2}-5x+2=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=2{;}~b=-5~e~c=2\\\\\Delta=(b)^{2}-4(a)(c)=(-5)^{2}-4(2)(2)=25-(16)=9\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-5)-\sqrt{9}}{2(2)}=\frac{5-3}{4}=\frac{2}{4}=0,5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-5)+\sqrt{9}}{2(2)}=\frac{5+3}{4}=\frac{8}{4}=2\\\\S=\{0,5;~2\}$