Solving Cubic Equations - Questions - Class Mathematics School Portal

Solving Cubic Equations - Questions

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Question 1

28654

Solve \(\left( {x + 1} \right)\left( {x + 2} \right)\left( {2x - 1} \right) = 0\)

Answer

\(x = -1,{\rm{ }}x = -2,{\rm{ }}x = \dfrac{1}{2}\)

Worked Solution

\begin{align}
& (x+1)(x+2)(2x-1)=0\\
&\begin{array}{rrr}
x+1=0,\;\;\,&x+2=0,\;\;\,&2x-1=0\;\,\\
\therefore x=-1,&x=-2,&x=\dfrac{1}{2}
\end{array}
\end{align}

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