F = 2x^2 - 3x - 2. Choose correct option | vedclass

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Question

$F=2 x^{2}-3 x-2=0$

$2 x^{2}-4 x+x-2=0$

$2 x(x-2)+1(x-2)=0$

$(x-2)(2 x+1)=0$

$d F=(4 x-3) d x$

$x=2 \Rightarrow \quad d F=(4 x 2-3) d x

Vedclass · Accepted Answer

$x = - 1/2$ is position of stable equilibrium

Vedclass · Answer

$F=2 x^{2}-3 x-2=0$

$2 x^{2}-4 x+x-2=0$

$2 x(x-2)+1(x-2)=0$

$(x-2)(2 x+1)=0$

$d F=(4 x-3) d x$

$x=2 \Rightarrow \quad d F=(4 x 2-3) d x \quad \Rightarrow \quad d f=5 d x$

$x=-1 / 2 \quad d f \quad\left(4 x\frac{-1}{2}-1\right) d x=-5d x$

$x=-1 / 2$

$F = 2x^2 - 3x - 2$. Choose correct option

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