anonymous
  • anonymous
Sove for H =( √3-2/2 +1 ) / (√3-2/2 +2 )
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
is it \[H=\frac{\frac{\sqrt{3}-2}{2}+1}{ \frac{\sqrt{3}-2}{2}+2}\]
anonymous
  • anonymous
yes it is
anonymous
  • anonymous
lets add using this rule \[\frac{ a }{ b }+\frac{ c }{ d }=\frac{ ad +bc}{ bd }\]

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anonymous
  • anonymous
alternativley easier way is to multiply by 2 both on numerator and denominator
anonymous
  • anonymous
can you tell me what you get when you do that
anonymous
  • anonymous
how can I multiply it ?
anonymous
  • anonymous
\[H=\frac{( \frac{\sqrt{3}-2}{2}+1)2 }{ (\frac{\sqrt{3}-2}{2}+2)2 }=\frac{\sqrt{3}-2+2 }{ \sqrt{3}-2+4 }=\frac{ \sqrt{3} }{ \sqrt{3}+2 }=\frac{ 3 }{ 3+2\sqrt{3} }\]
anonymous
  • anonymous
|dw:1371049368420:dw|
anonymous
  • anonymous
|dw:1371049579212:dw|
anonymous
  • anonymous
is it correct ?
anonymous
  • anonymous
other wise \[\frac{ \frac{\sqrt{3}-2}{2}+1 }{ \frac{\sqrt{3}-2}{2}+2 }=\frac{ \frac{\sqrt{3}-2+2*1}{2} }{ \frac{\sqrt{3}-2+2*2}{2} }=\text{ the 2's cancel} \implies \frac{ \sqrt{3} }{ \sqrt{3}+2 }\]
anonymous
  • anonymous
|dw:1371056930690:dw|
anonymous
  • anonymous
ohh yeah xD i forgot , xD
anonymous
  • anonymous
|dw:1371056973724:dw|
anonymous
  • anonymous
so the denominator is √3+2 ?
anonymous
  • anonymous
yes cos the twos cancel
anonymous
  • anonymous
so the answer for h = √3 / √3+2 :D
anonymous
  • anonymous
yes
anonymous
  • anonymous
you can simplify by multiplying both num and den by \[\sqrt{3}\]
anonymous
  • anonymous
\[H=\frac{3}{3+2\sqrt{3}}\]
anonymous
  • anonymous
oww owkie ! :D Thanks a lot ! :D but can't i simplify it by squaring it ?
anonymous
  • anonymous
|dw:1371050912591:dw|
anonymous
  • anonymous
you cannot "simplify" a number by squaring it, any more than you can "simplify 5 by saying \(5^2=25\)
anonymous
  • anonymous
you can, however, rationalize the denominator
anonymous
  • anonymous
oww okay thanyou :D
anonymous
  • anonymous
\[\frac{\sqrt3}{\sqrt3+2}\] \[=\frac{\sqrt3}{\sqrt3+2}\times \frac{\sqrt3-2}{\sqrt3-2}\] \[=\frac{3-2\sqrt3}{3-2}=3-2\sqrt3\]
anonymous
  • anonymous
mistake there
anonymous
  • anonymous
\[=\frac{3-2\sqrt3}{3-4}=\frac{3-2\sqrt3}{-1}=2\sqrt3-2\]
anonymous
  • anonymous
huh where ? idont get it , why do you have to multiply it to \[\sqrt{3}-2\]
anonymous
  • anonymous
to get the radical out of the denominator
anonymous
  • anonymous
ohh . Geez thanks xD
anonymous
  • anonymous
since \((a+b)(a-b)=a^2-b^2\) if you have \[\sqrt3+2\] you can get rid of the radical by multiplying by \(\sqrt3-2\) which gives you \[(\sqrt3+2)(\sqrt3-2)=\sqrt{3}^2-4=3-4=-1\]
anonymous
  • anonymous
of course you also have to multiply the numerator by the same thing this usually comes under the heading of "rationalize the denominator" or "write in simplest radical form"

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