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

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 :)

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)`

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`