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\({|x|} = x^2 + x - 3\) and how would you solve it?

 May 6, 2018
edited by Guest  May 6, 2018
 #1
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To solve the equation for x....

 

\(|x|=x^2+x-3\\~\\ x=\pm(x^2+x-3)\\~\\ \begin{array}\ x=+(x^2+x-3)&\qquad\text{or}\qquad& x=-(x^2+x-3)\\~\\ x=x^2+x-3&&x=-x^2-x+3\\~\\ 0=x^2-3&&0=-x^2-2x+3\\~\\ 3=x^2&&x^2+2x-3=0\\~\\ \pm\sqrt3=x&&(x+3)(x-1)=0\\~\\ x=\sqrt3\quad\text{or}\quad x=-\sqrt3&\qquad\text{or}& x=-3\quad\text{or}\quad x=1 \end{array} \)

 

Now we need to test each possible solution to see if it makes the given equation true:

 

\(|\sqrt3|\,\stackrel?=\,\sqrt3^2+\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,3+\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,\sqrt3\)

 

Yes, so  \(x=\sqrt3\)  is a solution.

 

\(|-\sqrt3|\,\stackrel?=\,(-\sqrt3)^2-\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,3-\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,-\sqrt3 \)

 

No, so  \(x=-\sqrt3\)  is not a solution.

 

\(|-3|\,\stackrel?=\,(-3)^2-3-3\\~\\ 3\,\stackrel?=\,9-3-3\\~\\ 3\,\stackrel?=\,3\)

 

Yes, so  \(x=-3\)  is a solution.

 

\(|1|\,\stackrel?=\,1^2+1-3\\~\\ 1\,\stackrel?=\,1+1-3\\~\\ 1\,\stackrel?=\,-1\)

 

No, so  \(x=1\)  is not a solution.

 

So the solutions are  \(x=\sqrt3\)  and  \(x=-3\)  .

 May 6, 2018

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