solve for x
x = x
\[8\sqrt{2x-5}=7-2\sqrt{2x-5}\]
Combine similar terms.
I keep getting a crazy answer and I dont think its right. this is what I keep getting\[x= \left( 249 \over 200 \right)\]
add \(2\sqrt{2x-5}\) to both sides and start with \[10\sqrt{2x-5}=7\] then square both sides
\[100(2x-5)=49\] etc
your answer is close, but not quite
not \(\frac{249}{200}\) but \(\frac{549}{200}\)
I suggest dividing both sides by sqrt(2x-5) first, that makes it easier to simplify since both sides share the same square root expression. So let's do that: First divide both side b y sqrt(2x-5):\[\frac{8\cancel{\sqrt{2x-5}}}{\cancel{\sqrt{2x-5}}}=\frac{7}{\sqrt{2x-5}}-\frac{2\cancel{\sqrt{2x-5}}}{\cancel{\sqrt{2x-5}}}\]Add 2 to both sides. \[8=\frac{7}{\sqrt{2x-5}}-2\]\[10=\frac{7}{\sqrt{2x-5}}\]Square both sides.\[100=\frac{49}{2x-5}\]Now rearrange and solve for x.\[2x-5=\frac{ 49 }{ 100 }\] Can you solve for x from here? @theluckyistduck
oh np that was actually a 5 but I typed it in wrong it just looked really wrong. I didnt think it should be that odd.
yeah it is right, i thought maybe that was a typo on your part when i saw it was off by one digit
ya thanks
yw
bad bois
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