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Given f(x) and g(x), find all values of x for which f(x) = g(x). f(x)=x^2-6 and g(x)=10x-28 x=?? simplify, using radicals as needed.
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x^2 - 6 = 10x - 28 solve for x
\[x^2-10x=-22\] \[(x-5)^2=-22+25=3\] \[x-5=\pm\sqrt{3}\] \[x=5\pm\sqrt{3}\]
well, solving for x gives me (x^2+22)/10, but that's not the answer. It tells me: "Set x^2-6 equal to 10x-28 and solve for x. Begin by completing the quare and then use the square root property to get all solutions of x." I don't know how to do that... :(
Oh, thankssatellite :)
you get what i wrote i hope
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