Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
If |x+1|=2|x-1|, x=?
1) x<1
2) x>0
(I gave some time to other people present an alternate solution... it didn´t happen.)
Let´s offer an alternate approach:
$$? = x$$
$$\left. \matrix{
{\left| {x + 1} \right|^{\,2}} = {\left( {x + 1} \right)^{\,2}} \hfill \cr
{\left| {x - 1} \right|^{\,2}} = {\left( {x - 1} \right)^{\,2}}\,\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{equation}}\,\,{\rm{given}}}^{{\rm{squaring}}\,\,{\rm{the}}} \,\,\,\,\,\,{\left( {x + 1} \right)^2} = 4{\left( {x - 1} \right)^2}\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\, \Rightarrow \,\,\,\,3{x^2} - 10x + 3 = 0$$
$$3{x^2} - 10x + 3 = 0\,\,\,\,\,\mathop \Rightarrow \limits_{{\text{product = 1}}}^{{\text{sum}}\,{\text{ = }}\frac{{{\text{10}}}}{3}} \,\,\,\,\,\boxed{\,\,\,x = \frac{1}{3}\,\,\,\,{\text{or}}\,\,\,x = 3\,\,\,\,}\,\,\,\left( * \right)\,\,\,\,\,$$
Important: whenever an equation is squared - or put to any positive EVEN power - solutions are never lost, but new ones may be (undesirably) created.
That´s why all roots obtained (after the procedure) must be checked ("tested") in the original equation. Both values are solutions to the original equation.
$$\left( 1 \right)\,\,x < 1\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,x = \frac{1}{3}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.$$
$$\left( 2 \right)\,\,x > 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,x = \frac{1}{3}\,\,\,\,{\text{or}}\,\,\,x = 3\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
