Question

A maior raiz da equação -2x^2+3x+5=0 vale: *


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Answer 1

Resposta:

A maior raiz vale 5/2

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~-2x^{2}+3x+5=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=-2{;}~b=3~e~c=5\\\\\Delta=(b)^{2}-4(a)(c)=(3)^{2}-4(-2)(5)=9-(-40)=49\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(3)-\sqrt{49}}{2(-2)}=\frac{-3-7}{-4}=\frac{-10}{-4}=\frac{5}{2}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(3)+\sqrt{49}}{2(-2)}=\frac{-3+7}{-4}=\frac{4}{-4}=-1\\\\S=\{\frac{5}{2},~-1\}$

Answer 2

Explicação passo-a-passo:

$\sf -2x^2+3x+5=0$

$\sf \Delta=3^2-4\cdot(-2)\cdot5$

$\sf \Delta=9+40$

$\sf \Delta=49$

$\sf x=\dfrac{-3\pm\sqrt{49}}{2\cdot(-2)}=\dfrac{-3\pm7}{-4}$

$\sf x'=\dfrac{-3+7}{-4}~\Rightarrow~x'=\dfrac{4}{-4}~\Rightarrow~x'=-1$

$\sf x"=\dfrac{-3-7}{-4}~\Rightarrow~x"=\dfrac{-10}{-4}~\Rightarrow~x"=2,5$

A maior raiz é $\sf 2,5$