Help Me, please.

Simplify and Reduce these Radicals

{2-√3} / {2+√3}

ManuelBautista2019 Feb 26, 2018

edited by
ManuelBautista2019
Feb 26, 2018

edited by ManuelBautista2019 Feb 26, 2018

edited by ManuelBautista2019 Feb 26, 2018

edited by ManuelBautista2019 Feb 26, 2018

edited by ManuelBautista2019 Feb 26, 2018

#2**+1 **

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

.Melody Feb 26, 2018

#3**0 **

(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.

Guest Feb 27, 2018

#4**+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. |

TheXSquaredFactor Feb 27, 2018