Rationalizing (1 Viewer)

[cUfffsxd]

If x = 3 - 2sqrt(3) / 3 + 2sqrt(2), find the vaile of x + 1/x

thx for anyhelp

 

[life92]

This will become quite messy...
Are you sure the question wasn't
x = 3 - 2sqrt(3) / 3 + 2sqrt(3)
or x = 3 - 2sqrt(2) / 3 + 2sqrt(2)

But anyway, using the original x = 3 - 2sqrt(3) / 3 + 2sqrt(2)

x + 1/x = [(3 - 2sqrt3) / (3 + 2sqrt2)] + [(3+2sqrt2)/(3-2sqrt3)]
= [(3-2sqrt3)^2+(3+2sqrt2)^2] / [(3+2sqrt2)(3-2sqrt3)]
= [ 9 - 12sqrt3 + 12 + 9 + 12sqrt2 + 8] / [(3+2sqrt2)(3-2sqrt3)]
= [38 - 12sqrt3 + 12sqrt2] / [(3+2sqrt2)(3-2sqrt3)]

From here, you'd need to multiply top and bottom by (3-2sqrt2)(3+2sqrt3) which will rationalise the denominator.
The denominator will become -3 but the numerator..
(9 + 6sqrt3 - 6sqrt2 - 4sqrt6)(38 - 12sqrt3 + 12sqrt2)
Yes, quite messy...

 

[fullonoob]

\frac{3-2\sqrt3}{3+2\sqrt2} + \frac{3+2\sqrt2}{3-2\sqrt3}

= 9-6(\sqrt2+\sqrt3) + 4 \sqrt6+\frac{9+6(\sqrt3+\sqrt2) + 4\sqrt6}{-3}


i swear equation editor is hard to use!!! i cant freaken find the signs LOL
Rationalise them both individually
then simplify
just common denominator after that cancel out the like terms
...how do you enter on equation editor

 

[fullonoob]

 

[cUfffsxd]

Life92 you were right o: x = 3 - 2sqrt(2) / 3 + 2sqrt(2) is the right equation.. sry ): was a bit tired. But thx guys. I'm sort of getting it i think XD, the fraction in a fraction part throws me off (1/x). But thx

 

[life92]

Haha thats allright.

Since you have x as a fraction, x = a/b, 1/x just equals the reciprocal, which is b/a.

So if x = 3 - 2sqrt(2) / 3 + 2sqrt(2)

x + 1/x = [ (3 - 2sqrt2) / (3 + 2sqrt2) ] + [ (3 + 2sqrt2) / (3 - 2sqrt2) ]
= [(3-2sqrt2)^2 + (3 + 2sqrt2)^2 ] / [ (3 + 2sqrt2)(3 - 2sqrt2) ]
= [ 9 - 12sqrt2 + 8 + 9 + 12sqrt2 + 8 ] / 1
= 34

For this type of question, it really depends on the numbers on what you should do first.

Here, either adding them or rationalising them first, both will arrive at the answer fairly quickly because it simplifies nicely