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Help Me, please.

Simplify and Reduce these Radicals

{2-√3} / {2+√3}

 Feb 26, 2018
edited by ManuelBautista2019  Feb 26, 2018
edited by ManuelBautista2019  Feb 26, 2018
edited by ManuelBautista2019  Feb 26, 2018
 #1
avatar+578 
+1

7-4√3

 

8)

 Feb 26, 2018
 #2
avatar+107030 
+1

\(2-\frac{\sqrt3}{2}+\sqrt3\\ =2\qquad +\sqrt3-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3}{2}-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3-1\sqrt3}{2}\\ =2\qquad +\frac{1\sqrt3}{2}\\ =2+\frac{\sqrt3}{2}\\\)

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 Feb 26, 2018
 #3
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(2-3^(1/2))/(2+3^(1/2)) or 2-3^1/2

                                         2+3^1/2

 

3^1/2 cancels and you are left with 2/2 which equals one.

 

 

^ is the same thing as an exponent.

 

and to clarify 3^1/2, that is the same as saying the square root of 3.

 Feb 27, 2018
 #4
avatar+2345 
+1

It appears as if this question has been edited since the replies of other answerers. As of now, the expression to simplify is \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\)

 

\(\frac{2-\sqrt{3}}{2+\sqrt{3}}\) Radicals in the denominator are considered to be improper; it is required that all radical terms are moved to the numerator. This can be achieved by utilizing the conjugate of the denominator: \(2-\sqrt{3}\)
\(\frac{2-\sqrt{3}}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}\) To simplify the denominator, it is possible to use the algebraic rule that \((a+b)(a-b)=a^2-b^2\). Notice that this process does not change the inherent value of the expression since all I am doing is multiplying by 1, which has no effect.
\(\frac{\left(2-\sqrt{3}\right)^2}{2^2-\sqrt{3}^2}\) By squaring the radical in the denominator, we are getting rid of a radical's existence from the denominator. 
\(\frac{\left(2-\sqrt{3}\right)^2}{4-3}\) Well, it is very clear at this point that the denominator simplifies to 1. Expand the numerator using the rule that \((a-b)^2=a^2-2ab+b^2\)
\(2^2-2*2*\sqrt{3}+\sqrt{3}^2\) Simplify.
\(4-4\sqrt{3}+3\)  
\(7-4\sqrt{3}\) This is what OfficialBubbleTanks got.
   
 Feb 27, 2018
 #5
avatar+1451 
+1

Thanks To all of you!!!laugh

 Feb 27, 2018

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