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2015-04-14T15:06:58+05:30

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3-\sqrt{\dfrac{17}{4}+3\sqrt{2}}\\3-\frac{1}{2}\sqrt{17+12\sqrt{2}}

Next try to write 17+12\sqrt{2} as a perfect square using the identity
a^2+b^2+2ab = (a+b)^2

with some trial and error you will be able to break like this
17+12\sqrt{2} = 9+8+2*6*\sqrt{2} = 3^2 + \sqrt{8}^2 +2*3*\sqrt{8} = (3+\sqrt{8})^2

plug this in your main expression and simplify
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awesome thanks
yw! the trick is to express 17+12sqrt(2) as a perfect square using that identity
yup superb thanku
2015-04-15T04:59:04+05:30

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If you want to find such solutions to square roots of some irrational numbers, in general we can follow the following procedure.  We arrive at a quadratic equation.  We find the solution and again combine back to get the final answer.

if you are able to get a short cut easily, do that way,  Otherwise follow this way.

let\ x=\sqrt{\frac{17}{4}+3\sqrt2}\\\\x^2=\frac{17}{4}+3\sqrt2,\ \ \ \ 4x^2=17+12\sqrt2\\\\let\ 2x=a+b\sqrt2,\ \ where\ a\ b\ are\ u=real\ rational\ numbers\\\\squaring\ both\ sides, \ \ 4x^2=a^2+2b^2+2\sqrt2\ ab\\\\comparing\ the\ equations:\\a^2+2b^2=17, \ \ \ \ 2ab=12,\ \ \ ab=6,\ \ \ \ b=6/a\\\\a^2+2(6/a)^2=17\\a^4+72-17a^2=0\\\\a^2=[17+-\sqrt{289-288}]/2\\\\a^2=8\ or\ 9\\a=+2\sqrt2\ or\ -2\sqrt2\ \ or\ 3\ or\ \ -3

b=6/a=\frac{3}{\sqrt2},\ \ or\ -\frac{3}{\sqrt2},\ \ or\ 2\ \ or\ \ -2\\\\so\ 2x=2\sqrt2+3\ \ or\ \ -2\sqrt2-3\ \ or\ \ 3+2\sqrt2\ or\ -3-2\sqrt2\\\\answer\ is\ \ 3-x=3-\frac{3+2\sqrt2}{2}=1.5-\sqrt2

Here we have only taken the positive values of a and b.

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