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burhan101
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\[\huge 0=1- \frac{ 1 }{ 2x^{\frac{ 3 }{ 2}}} \]
Abhishek619
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|dw:1371705222820:dw|
burhan101
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how does it equal one ?!
terenzreignz
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slowly :D
Why don't we bring \(\huge \frac1{2x^{\frac32}}\)
to the left side like this...
\[\Large \frac1{2x^{\frac32}}=1\]
clear now? :)
burhan101
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|dw:1371743475441:dw|
burhan101
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yes
terenzreignz
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We basically added \[\huge \frac1{2x^{\frac32}}\]
to both sides as per the addition property of equality.
terenzreignz
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Well then, can you solve it from here?
burhan101
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do i multiply by both sides?
terenzreignz
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Sure :)
burhan101
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|dw:1371743682268:dw|
terenzreignz
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Never mind that for now. Why don't we multiply both sides by
\[\Large 2x^{\frac32}\]
That gives us...
\[\Large 1 = 2x^{\frac32}\]
And divide both sides by 2...
burhan101
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\[x^{3/2} =1/2 \]
terenzreignz
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true dat :)
Now you can raise both sides to the \(\LARGE \frac23\) power (just to get rid of the exponent of x)
You then get...?
burhan101
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|dw:1371743929750:dw| like this?
terenzreignz
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That gives you x.
What about the right-hand side?
burhan101
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0/63
burhan101
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0.63 *
terenzreignz
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come again? :D
terenzreignz
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Oh, were you supposed to give it in decimal form? okay then :D
burhan101
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I can get a fraction answer too?
terenzreignz
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Well, the best you can hope for is the ugly-looking
\[\Large \left(\frac12\right)^{\frac23}\]
terenzreignz
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and that is the correct answer :)
burhan101
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oh so i should probably keep it,
burhan101
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thank youuu !!!
terenzreignz
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Yeah, unless you were specifically asked for an approximation :)
terenzreignz
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Oh well... now I have to go :)
Please practice solving algebraic equations :D
-----------------------------------------
Terence out