+1495 Help Me, please.
Simplify and Reduce these Radicals
{2-√3} / {2+√3}
+118687 \(2-\frac{\sqrt3}{2}+\sqrt3\\ =2\qquad +\sqrt3-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3}{2}-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3-1\sqrt3}{2}\\ =2\qquad +\frac{1\sqrt3}{2}\\ =2+\frac{\sqrt3}{2}\\\)
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(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.
+2446 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. |
