Lê Quốc Trần Anh Coordinator
03/01/2018 at 16:54-
Alone 03/01/2018 at 17:01
We have:\(64x+16x^2-4x^3-x^4=0\)
\(\Leftrightarrow16x\left(x+4\right)-x^3\left(x+4\right)=0\)
\(\Leftrightarrow\left(16x-x^3\right)\left(x+4\right)=0\)
\(\Leftrightarrow x\left(x+4\right)\left(16-x^2\right)=0\)
\(\Leftrightarrow x\left(4-x\right)\left(x+4\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\4-x=0\\x+4=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Lê Quốc Trần Anh selected this answer. -
FA Liên Quân Garena 08/01/2018 at 21:50
We have:64x+16x2−4x3−x4=0
⇔16x(x+4)−x3(x+4)=0
⇔(16x−x3)(x+4)=0
⇔x(x+4)(16−x2)=0
⇔x(4−x)(x+4)2=0
⇒⎡⎢⎣x=04−x=0x+4=0
⇒⎡⎢⎣x=0x=4x=−4