How do you simplify #(sqrt5-3)^2#?

3 Answers
Nov 20, 2017

Answer:

#14 - 6sqrt5#

Explanation:

Recall the identity for a binomial squared:

#(a+b)^2 = a^2 + b^2 + 2ab#

So, all your doing is using this formula, but with two numbers. This gives you:

#(sqrt5-3)^2 = (sqrt(5))^2 + (-3)^2 + 2(sqrt(5)(-3))#

#= 5 + 9 - 6sqrt(5)#

# => 14 - 6sqrt5#

However, there is also another formula you might stumble across that you could use:

#(a-b)^2 = a^2 + b^2 - 2ab#

Can you use this and get the same answer? YES! This is exactly the same thing . Instead of plugging a negative number into the first formula, we just take care of that by making #b# negative beforehand, and then dealing purely in positive numbers.

Either way, you should end up at the same answer.

Hope that helped :)

Nov 20, 2017

Answer:

#= 2(7-3sqrt5)#

Explanation:

Use identity: #(x-y)^2 = x^2- 2xy+y^2#

#(sqrt5-3)^2#----- here #x= sqrt5 and y = 3#

#=>(sqrt5-3)^2= (sqrt5)^2 - 2xxsqrt5xx3 +(3)^2#

#=>(sqrt5-3)^2= 5 - 6sqrt5 +9#

#=> (sqrt5-3)^2= 14 - 6sqrt5 #

#(sqrt5-3)^2= 2(7-3sqrt5)#

Nov 20, 2017

Answer:

#5 - 6sqrt(5) + 9#

Explanation:

First, let #sqrt(5)# be #a# and #-3# be b.

Our simplified answer will be in the form of #a^2 +2ab + b^2#.

First, to get #a^2#, we do this: #sqrt(5)^2# means that the #sqrt# and squared get cancelled. So that becomes #5#.

Then we need #2ab#. #2(-3)(sqrt(5)) = -6sqrt(5)#

Finally, we need #b^2#, which is #(-3)^2 -> 9#

So let's put all of these together, so our final answer is:
#5 - 6sqrt(5) + 9#