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C

congchuaanhsang

$x^2+4Rx+R^2$=0

$\Delta$=$16R^2+4R^2$=$20R^2$\Rightarrow $\sqrt{ \Delta }$ =$2\sqrt{5}R$

$x_1$=$\dfrac{-4R-2\sqrt{5}R}{2}$=$\dfrac{(-4-2\sqrt{5})R}{2}$

$x_2$=$\dfrac{(-4+2\sqrt{5})R}{2}$
 
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