I've been trying to complete this question for a while now, but my answers never make sense. I always end up with $3x = x -1$, or something similar, but that can't possible to right. The instructions are 'solve each of the following radical equations algebraically. State any restrictions on the variable.' Any pointers in the right direction would be greatly appreciated. Thank you in advance!
$\sqrt{3x+5}-\sqrt{x-1}=0$
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algebra-precalculus
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0How do you end up with $3x=x-1$? – 2017-01-12
2 Answers
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if it is $$\sqrt{3x+5}-\sqrt{x-1}=0$$ we can write $$\sqrt{3x+5}=\sqrt{x-1}$$ after squaring we have$$3x+5=x-1$$ thus $$x=-3$$ since we have $$x\geq 1$$ we have no solutions.
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Hint
$$\sqrt{3x+5}-\sqrt{x-1}=0\rightarrow \sqrt{3x+5}=\sqrt{x-1}$$
Square both sides and remember that
$$3x+5 \ge 0\\ x-1\ge 0 $$