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Solve for $x$ in the given equation. $\dfrac{\sqrt{x}}{\sqrt{3}x+\sqrt{2}} = \dfrac{1}{2\sqrt{6}x+4}$

 Dec 18, 2014

Best Answer 

 #1
avatar+23254 
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Multiply the left side by:  2√2 / 2√2:

Left side becomes:  [√x / (√3x + √2)] · [2√2 / 2√2]  =  2√(2x) / [ 2√6x + 4 ]

Now the left side and the right side have the same denominators. For the two sides to be equal, they must have equal numerators:    2√(2x)  =  1  

--->                            √(2x)  =  1/2                 divide by 2

--->                                 2x  =  1/4                  square both sides

--->                                   x  =  1/8                  divide by 2

 Dec 18, 2014
 #1
avatar+23254 
+5
Best Answer

Multiply the left side by:  2√2 / 2√2:

Left side becomes:  [√x / (√3x + √2)] · [2√2 / 2√2]  =  2√(2x) / [ 2√6x + 4 ]

Now the left side and the right side have the same denominators. For the two sides to be equal, they must have equal numerators:    2√(2x)  =  1  

--->                            √(2x)  =  1/2                 divide by 2

--->                                 2x  =  1/4                  square both sides

--->                                   x  =  1/8                  divide by 2

geno3141 Dec 18, 2014

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