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What quadratic equation has solutions 2 and -1/3 use the variable X to write your equation

User Yazu
by
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1 Answer

3 votes

Given the roots,

2 and -1/3

Therefore we can write,


(x-2)(x+(1)/(3))=0

On solving,


\begin{gathered} (x-2)(x+(1)/(3))=0 \\ x^2+(1)/(3)x-2x-(2)/(3)=0 \\ x^2+(x-6x)/(3)-(2)/(3)=0 \\ 3x^2-5x-2=0 \end{gathered}

User Qqtf
by
8.2k points

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