What I have done is squared both sides which gives x=(6-x)^2 = 0 = x^2-13x+36 0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function. Is there a better way of doing this??
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Originally Posted by p75213 What I have done is squared both sides which gives x=(6-x)^2 = 0 = x^2-13x+36 0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function. Is there a better way of doing this?? So or . Therefore or . By the way, both values of fit the function...
I see your point. Thanks for that.
Last edited by p75213; May 30th 2010 at 10:45 PM.
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