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1 The problem is to find all real solutions (if any exists) for $sqrt{2x-3} +x=3$ . Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them: I got rid of the root by squaring both sides, $$sqrt{2x-3}^2=(3-x)^2$$ $$0=12-8x+x^2$$ Using the AC method, I got $$(-x^2+6x)(2x-12)=0$$ $$-x(x-6)2(x-6)=0$$ $$(-x+2)(x-6)=0$$ hence, $$x=2, x=6$$ Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach? algebra-precalculus share | cite | improve this question