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How do you rationalize the denominator #2/(5-sqrt3)#?

Answer 1

#2/(5-sqrt3)#

We can ratioinalise the denominator by multiplying the numerator and denominator of the expression by the conjugate of #5-sqrt3#

The conjugate of #5-sqrt3 = color(red)(5+sqrt3#

#=(2xxcolor(red)((5+sqrt3)))/((5-sqrt3)xxcolor(red)((5+sqrt3))#

We know that #color(blue)((a-b)(a+b)=a^2 - b^2 # so ,#(5-sqrt3)xx(5+sqrt3) = 5^2- sqrt3^2 = color(blue)(25 - 3 = 22#

the expression becomes: #(cancel2xxcolor(red)((5+sqrt3)))/cancel22#

# = (5+ sqrt3)/ 11#

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Answer 2

Eva Abramson

To rationalize the denominator 2/(5-sqrt3), we multiply both the numerator and denominator by the conjugate of the denominator, which is 5+sqrt3. This results in (2 * (5+sqrt3)) / ((5-sqrt3) * (5+sqrt3)). Simplifying further, we get (10+2sqrt3) / (25-3). The final rationalized form is (10+2sqrt3) / 22.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.