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References External links Category:Software companies of CroatiaQ: How to solve the cubic equation $x^3-2x^2+7x-3=0$?
How can this cubic equation: $$x^3-2x^2+7x-3=0$$ be solved? I know that when you encounter a cubic equation, there arethree cases, but I can't see how to apply them here. Thanks. A: Consider the solutions to the quadratic $$ (x-1)(x-3)=0 $$ Thesolutions are $$ x=1, x=3 $$ The equation can be written as $$ (x-1)(x-3) + \epsilon x^2 = 0 $$ The discriminant $\Delta =(-3)^2 + 4 \cdot 1 \cdot (-1) = 7$ is never $0$ and so we need to consider the zeros to the cubic $$ f(x) = x^3 - 2x^2 + 7x-3 $$At this point you have three cases... Case 1) If $x=1$, we get $0 = 3 - 2 \cdot 1 + 7 - 3$ This adds up to $0 = -0 + 0 + 0 - 0 = 0$.
The roots are all real. Case 2) If $x=3$, we get $0 = 9 - 6 \cdot 3 + 21 - 9$. This adds up to $0 = 0 - 18 + 0$. The roots are all
real. Case 3) If $\Delta > 0$, the cubic has real roots. So if we have $$ x^3 - 2x^2 + 7x-3 = 0 $$ We will have real roots if wehave $-2 0$, or equivalently $$ 1 0$ or $$ 7 > 4 \cdot 1 \cdot -2 + 7 > 4 \cdot 1 \cdot (-2 + 7) $$ or equivalently $$ 0 > -9 + 12
\cdot 1 > 0 $$ and this holds only if $1 f678ea9f9e
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OEM WIFI CarBrain C168 Bluetooth Scanner Free Download