\({|x|} = x^2 + x - 3\) and how would you solve it?
+9466 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 \(x=-3\) .
