If x = 2+square root of 3 divided by 2-square root of 3, find: x-1 over x
For messy problems like this, I like to substitute \[a = \sqrt{3}\]do the algebra, then substitute back at the end, after the dust has settled. Your mileage may vary.
\[x=(2+\sqrt{3})/(2-\sqrt{3})\] multiply conjugate -> 2+sqrt(3) to top and bottom \[=(2+4\sqrt{3}+3)\] plugin for x-1/x \[((2+4\sqrt{3}+3)-1)/(2+4\sqrt{3}+3) =( 4+4\sqrt{3})/(5+4\sqrt{3})\] multiply conjugate again -> 5-4sqrt(3) to top and bottom = \[(4\sqrt{3}-28)/53\] = aprox. -0.398
the answer is 8 multiply square root of 3
Well, can't shoot a guy for trying. It's been a while
thanks for trying
simplifying, you should get \(\large x=7+4\sqrt3 \)
now plug that in for (x-1)/x
and simplify again....
\[x=\frac{ (2+\sqrt{3})}{(2-\sqrt{3}) } \times \frac{ (2+\sqrt{3}) }{(2+\sqrt{3}) } = \] You'll get what @dpaInc posted.
right i know that part but then i was stuck on the second part. i didnt know i had to plug it
\[\frac{ x -1 }{ x } =\frac{ 7+4\sqrt3-1 }{ 7+4\sqrt3 } \times \frac{ 7-4\sqrt3 }{ 7-4\sqrt3 }\]
^ again, I multiplied by the conjugate of the denominator \[\frac{ 6+4\sqrt3 }{ 7+4\sqrt3 } \times \frac{ 7-4\sqrt3 }{ 7-4\sqrt3 } = \frac{ 42+28\sqrt3-24\sqrt3-16*3 }{49-16*3 }\]
i have the answer as 8 multiplied by square root of 3
hmm... maybe I've made a mistake.
Nope, my answer is correct.
i have the answer in my textbook as 8 multiplied by square root of 3
Are you sure you wrote the problem correctly? is it \[x=\frac{ (2+\sqrt{3})}{(2-\sqrt{3}) } \]
Check it on a calculator. x = 13.928 Then find\[\frac{ x-1 }{ x }\] and you'll get 0.928. Then check \[8\sqrt3 = \] and you'll get 13.856.
ill check it just give me a sec
they are almost the same. im wondering where 8 multiplied by square root of 3 came from
They aren't almost the same. x is close to 8sqrt3... but (x-1)/x is not even close to 8sqrt3.
i believe this problem can only be solved by a calculator as it is a digital worksheet. sorry made a mistake
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