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How do you rationalize the denominator and simplify #sqrt(10)/(sqrt(5)-2)#?

Answer 1

Kennedy Adams

You first multiply top and bottom with #sqrt5+2#

This will give the special product #A^2-B^2# as denominator

#=sqrt10/(sqrt5-2)*(sqrt5+2)/(sqrt5+2)#

#=(sqrt(2*5)*sqrt5+2sqrt10)/(sqrt5^2-2^2)#

#=(sqrt(2*5^2)+2sqrt10)/(5-4)#

#=5sqrt2+2sqrt10#

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Answer 2

Nathan Axel

To rationalize the denominator and simplify the expression sqrt(10)/(sqrt(5)-2), we can multiply both the numerator and denominator by the conjugate of the denominator, which is sqrt(5)+2. This will eliminate the square root in the denominator.

By applying the conjugate, we get: (sqrt(10)/(sqrt(5)-2)) * ((sqrt(5)+2)/(sqrt(5)+2))

Simplifying this expression, we have: (sqrt(10)(sqrt(5)+2))/((sqrt(5)-2)(sqrt(5)+2))

Expanding the denominator using the difference of squares, we get: (sqrt(10)*(sqrt(5)+2))/(5-4)

Further simplifying, we have: (sqrt(10)*(sqrt(5)+2))/1

Finally, we can simplify the expression to: sqrt(10)*(sqrt(5)+2)

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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.