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3 years ago

What are the real or imaginary solutions of the polynomial x^4-3x^2=-2x

Mathematics

2 answers:

SpyIntel [72]3 years ago

5 0

x⁴-3x²= -2x

Step-by-step:

x⁴-3x²+2x=0

x*(x³-3x+²)=0

x*(x³-4x+x+2)=0

x*(x*(x²-4)+x+2)=0

x*(x*(x-2)*(x+2)+1)=0

x*(x+2)*(x*(x-2)+1)=0

x*(x+2)*(x²-2x+1)=0

Using a²-2ab + b²=(a-b)², factor the expression.

x*(x+2)*(x-1)²=0

Therefore x=0

Therefore x+2=0

(X-1)²=0

X=0

X=-2

X=1

Therefore

x1 is -2

x2 is 0

x3 is 1.

bagirrra123 [75]3 years ago

4 0

Answer:

x = 0 or 1 or -2

Step-by-step explanation:

$x^{4} -3x^{2} +2x=0$

x(x^3-3x+2)=0

x(x-1)^2 * (x+2)0

so x = 0 or 1 or -2

there is no imaginary solutions

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3 years ago

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labwork [276]

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Step-by-step explanation:

because I did this

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Which are the like terms in 2y^3 – 4y^2 + y + y^3?

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$\bf \begin{array}{llll}
\cfrac{(x-{{ h}})^2}{{{ a}}^2}+\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1\\\\ \cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1
\end{array}\quad 
\begin{cases}
\textit{major axis}=a+a\\
a=\textit{the larger denominator}\\
b=\textit{smaller denominator}
\end{cases}\\\\
-----------------------------\\\\
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\begin{cases}
a=4\\
b=2
\end{cases}$

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3 years ago

What are the real or imaginary solutions of the polynomial x^4-3x^2=-2x​

What are the real or imaginary solutions of the polynomial x^4-3x^2=-2x