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 July 9th, 2008, 04:53 PM #1 Newbie   Joined: Jul 2008 Posts: 1 Thanks: 0 NEED Help Quick i need quick help to solve these problem i keep running into problems when i check my answer Square root of 3x+1 = 3+ square root of x-4 and absolute value of 2x+1 = x-1
 July 9th, 2008, 10:15 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: NEED Help Quick Hi $\sqrt{3x+1}=3+\sqrt{x-4},\hspace{60mm}(\ast)$ which we rearrange to $\sqrt{3x+1}-\sqrt{x-4}=3,$ then square both sides: $(3x+1)+(x-4)-2\sqrt{(3x+1)(x-4)}=9.$ Then we have $4x-12=2\sqrt{3x^2-11x-4},$ so: $(2x-6)^2=3x^2-11x-4\\ 4x^2-24x+36=3x^2-11x-4\\ x^2-13x+40=0\\ (x-(x-5)=0." /> The two solutions are $x=5$ and $x=8$ (which we can check are both consistent with $(\ast).$) Secondly, $|2x+1|=x-1$ can be examined in two separate cases: $x\geq -\frac12$ and $x<-\frac12.$ If $x\geq-\frac12,$ then $|2x+1|=2x+1,$ so $2x+1=x-1.$ This has one solution at $x=-2,$ but this is not valid since we are looking only at $x\geq-\frac12.$ On the other hand, if $x<-\frac12,$ then $|2x+1|=-1-2x,$ so $-1-2x=x-1.$ This has one solution at $x=0,$ but again this is not valid since we are looking only at $x<-\frac12.$ So $|2x+1|=x-1$ has no solutions.
 July 10th, 2008, 09:15 PM #3 Member   Joined: Feb 2008 Posts: 89 Thanks: 0 Re: NEED Help Quick [color=#4000BF]Dear Helpless: |2x + 1| = x - 1 is readily solved if you consider the graphs of y1 = |2x + 1| and y2 = x - 1. y2 has x-intercept at (1, 0), with slope = 1. y1 has x-int. at (-1/2, 0). Clearly, were there solutions, y2 would intersect the "right-most" branch of y1 in the first quadrant. Given the aforesaid, however, a moment's reflection indicates that no such intersection exists, i.e., the equation is without solutions. Regards, Rich B.[/color] rmath4u22aol.com

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