Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Radical Expressions Containing Variables - To simplify a radical expression with variables: --If the exponent is even, divide the exponent by two, and that will be your new exponent Ex. βπ₯ 12 = β(π₯ 6 )2 = π₯ 6 Ex. βπ 4 = β(π 2 )2 = π2 Ex. βπ₯ 24 = Ex. βπ16 = --If the exponent is odd, separate into two terms. The first term will be the exponent inside the radical sign. The second term will be the exponent one less divided by 2 Ex. βπ₯ 5 = βπ₯ 4 (π₯ 1 ) = π₯ 2 βπ₯ Ex. βπ11 = βπ10 (π1 ) = π5 βπ Ex. βπ15 = Ex. βπ₯ 23 = - If the square root of a variable raised to an even power has a variable raised to an odd power for an answer, the answer must have absolute value signs. This ensures that the answer will be positive. Ex. βπ₯ 6 = β(π₯ 3 )2 = |π₯ 3 | = |π₯ |3 Ex. β(β2)10 = β(β25 )2 = |(β2)5 | = 25 Ex. βπ₯ 14 = Ex. βπ 22 = Simplifying Non-Perfect Square Radicals - To simplify a radicand, first separate the number into two factors where one factor is a perfect square. - Then, simplify the perfect square and keep the second factor as a radical. If there is a radical, the number is considered irrational. - If after factoring you notice that you could have another perfect square, factor again until there are no perfect squares left! - If there is a number outside of a radical, multiply the number to any simplified perfect squares. Ex. β8 = β4β2 = 2β2 Ex. β125 = β25β5 = 5β5 Ex. β72 = Ex. β18 = Ex. β360 = Ex. β24 = Ex. β108 = Ex. 2β24 = Simplifying Roots of Variables - We will now combine everything we did on the first two lessons. - With variables, divide the exponent by 2. The number of times that 2 goes into the exponent becomes the power on the outside of the radical and the remainder is the power of the radicand. βπ₯ 7 = π₯ 3 βπ₯ Note: Absolute value signs are not needed because the radicand had an odd power to start. - With numbers, simplify by simplifying the perfect square out of the radicand and keeping the remaining factor inside the radical sign. - Combine both answers. Ex. β50π₯ 4 π¦12 π§ 3 = β25β2βπ₯ 4 βπ¦12 βπ§ 2 βπ§ = 5 β2 π₯ 2 π¦ 6 = 5π₯ 2 π¦ 6 π§β2π§ Ex. β8π₯ 5 π¦ 6 π§ 4 = π§ βπ§ Ex. β16π₯ 7 π¦ 4 π§ = Ex. βπ11 π22 = Ex. β40π14 π 7 = Ex. β100π₯ 3 π¦10 π§ 4 = Solving Equations with Perfect Squares and Cube Roots - The product of two βequalβ factors is the SQUARE of the number (5)(5) = 25 - The product of three βequalβ factors is the CUBE of the number (5)(5)(5) = 125 - When solving equations, you need to find the proper root based on what the question is asking. Note: If the root you are looking for is square, you must show a ± sign to show the root could be positive or negative! Note: If the CUBE is negative, then your root is also negative! Ex. 5π₯ 2 = 605 5π₯ 2 5 = 605 5 π₯ 2 = 121 βπ₯ 2 = β121 π₯ = ±11 Ex. π₯3 9 π₯3 9 = β3 (9) = β3(9) π₯ 3 = β27 3 βπ₯ 3 = ββ27 π₯ = β3 Ex. 4π₯ 3 = 500 Ex. π₯2 8 = 18