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how do i solve a a problem like this? plz explain in detail
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\[\sqrt{3x+4} = 1 + \sqrt{3x-11}\]
square both sides first
3x+4 = 1+3x-11 +2(3x-11)^1/2
but then how do I get rid of the radical so I can solve for x?
14=2(3x-11)^1/2 7= (3x-11)^1/2 squaring again we get
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49=3x-11 3x=60 x=20
oh I see...thanks!
good one !!
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