Square root??

You cannot have the denominator (bottom #) in square root.

In this case, get rid of the +√５ in (-2+√５), we will multiply the conjugate of (-2+√５) which is (-2-√５) to cancel out the square root.

(-2+√５)(-2-√５) >>> FOIL method.

4 + 2√５ – 2√５ – √25 >>> 2√５ – 2√５ = 0

4 – √25

4 – 5

-1

The denominator is now -1

Remember when you multiply the denominator (bottom) by something, you will have to multiply the same thing to the numerator (top)

(1+√５)(-2-√５)

-2 -√５- 2√５- 5 >>> -√５is the same thing as -1√5, rewrite

-2 – 1√５- 2√５- 5 >>> Combine like terms, – 1√５and – 2√５

-2 – 3√５ – 5 >>> Combine like terms, -2 and -5

-7 – 3√ 5

The numerator is now -7 – 3√ 5

Now you have -7 – 3√ 5 / -1

Multiply both top and bottom by -1 (or -1/1) to cancel out the -1 on the bottom of -7 – 3√ 5 / -1

Top: (-7 – 3√ 5)(-1) = 7 + 3√ 5

Bottom: (-1)(-1) = 1

7 + 3√ 5/1 is the same thing as 7 + 3√ 5.

you need to simplify sq. root 20 into 2 sq. root of five and sq. root of 8 into 2 sq. root of two thus, you need to element out 2, which leaves 2(sq. root of five + sq. root of two)^2 by potential of simplifying extra, 2(5+2square root of five * sq. root of two + 2) 2(7+ 2 sq. root of 10) thus the appropriate answer could be: 14+ 4square root of 10

multiply and divide by (1-√５) and then solve the rest..

( 1 + sqrt(5)) / (-2 + sqrt(5))

multiply the numerator and the denominator by (-2 – sqrt(5)) “the conjugate of the denominator:

numerator becomes: (1+sqrt(5)) * (-2-sqrt(5)) = 1*-2 – 1*sqrt(5)

-2*sqrt(5) – sqrt(5)*sqrt(5) = -2 -sqrt(5) -2 sqrt(5) – 5 =

-7 – 3sqrt(5)

denominator becomes: ( -2 + sqrt(5)) * ( -2 – sqrt(5)) =

-2*-2 + 2 * sqrt(5) – 2 * sqrt(5) – sqrt(5)*sqrt(5) = 4 – 5 = -1

numerator/denominator = ( -7 – 3sqrt(5)) /( -1) =

(-7)/(-1) – 3sqrt(5)/(-1) = 7 + 3 sqrt(5)

Thanks

i love rooting squares