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Solve... Radical 3x/ Radical 2x + 3=2
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\[\sqrt{3x}\div \sqrt{2x+3}=2\]
start with \[\frac{3x}{2x+3}=4\]
What would be my first steps to solve this please?
that is, square both sides
Square both sides of the equation! Got it!
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then \[3x=4(2x+3)\]
etc
don't forget to check at the end because you squared and that can mess things up. so check your solution make sure it is valid
I got 12/-5, is that correct?
yes but that means there is no solution
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why is there no solution?
because you cannot take the square root of a negative number. so don't say "the answer is \[-\frac{12}{5}\] say there is no solution
OK - Thanks!
yw
I have another one, I will post on the other side!
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