How do you rationalize the denominator and simplify #1/(sqrt5-3)#?

Answer 1

#1/(sqrt(5)-3)=-(sqrt(5)+3)/4#

For expressions #text(something)/(sqrt(a)-b)# we multiply the numerator and denominator by #sqrt(a)+b# so it matches the LHS of the formula #(x+y)(x-y)=x^2-y^2#.

#1/(sqrt(5)-3)=1/(sqrt(5)-3) * (sqrt(5)+3)/(sqrt(5)+3)=(sqrt(5)+3)/(5-9)=-(sqrt(5)+3)/4#

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Answer 2

Genesis Allen

To rationalize the denominator and simplify 1/(sqrt5-3), we multiply both the numerator and denominator by the conjugate of the denominator, which is sqrt5+3. This results in (1 * (sqrt5+3)) / ((sqrt5-3) * (sqrt5+3)). Simplifying further, we get (sqrt5+3) / ((sqrt5)^2 - (3)^2), which simplifies to (sqrt5+3) / (5-9), and finally, (sqrt5+3) / (-4).

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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.