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Dividing binomials with square roots?
(sqrt(2) + sqrt(5)) / (sqrt(2) - sqrt(5))
I forgot how to divide binomials with square roots in it.
2 Answers
- 8 years agoFavorite Answer
((sqrt(2) + sqrt(5))/(sqrt(2) - sqrt(5)))*((sqrt(2) + sqrt(5))/(sqrt(2) + sqrt(5))) ... common denominator
((sqrt(2) + sqrt(5))^2)/(2+sqrt(10)-sqrt(10)-5) ... simplifying top and bottom
(2+(2*sqrt(10))+5)/(-3) ... expanding top (a^2 + 2ab + b^2) and further simplifying bottom
7+(2*sqrt(10))/(-3) OR (-7/3)-((2*sqrt(10))/3) ... further simplifying top and keeping bottom
There are multiple ways to express the final answer, but I have just listed two. I tried to type it out neatly in Microsoft Word with math symbols, but it would not copy over. Hope it helps.
Source(s): My Background in Math - NiallLv 78 years ago
Rationalise the denominator by multiplying both the numerator and denominator by the denominator but with the opposite sign (√2 + √5) and foiling it out. Notice how the two surds cancel each other out:
(√2 + √5) (√2 + √5)
= 2 + √10 + √10 + 5
= 7 + 2√10
(√2 - √5) (√2 + √5)
= 2 + √10 - √10 - 5
= 2 - 5
= -3
(7 + 2√10) / (-3)
It's customary (but optional) to have the denominator positive:
-(7 + 2√10) / (3)
That is now in it's simplest form.
