Multiplying Square Roots with ExponentsEx: √32 x √44 32/2 x 44/2 31 x 42 3 x 16 48 Notice that our new power for the base of 3 became 1, and the new power for the base of 4 became 2. Hence, our problem became 3 multiplied by 16. Sometimes, our exponents do not divide evenly by our root of 2. When this happens, a base with the power of 1 will have to remain in the radical. To work these problems, we separate our base into two terms, one with a power that divides evenly by two, and one with a power of 1. Example: √53 x √27 √(52 x 51) x √(26 x 21) 52/2 √ 5 x 26/2√2 51 √ 5 x 23√2 5 √ 5 x 8√2 (5 x 8) √(5 x 2) 40 √ 10 Notice a √5 and a √2 had to stay in the radicals because their powers did not divide evenly by our root of two. Finally, we must be able to this with variables as well. Example: 2√(56z9) x 3√(37y3) 2√(56z8z1) x 3√(3631y2y1) 2(56/2)(z8/2)√(5z) x 3(36/2)(y2/2)√(3y) 2(125)(z4) √(5z) x 3(27)(y1) √(3y) 250 z4 √5z x 81y1 √3y (250 x 81)y1 z4 √( 5z x 3y) 20,250 y1z4 √15yz We had to break up numbers as well as variables in order to bring out as much as possible. Notice that after simplifying, we combined the two terms by multiplying the coefficients together, as well as our bases. Practice Problems Simplify. 1. √64 x √38 2. √73 x √35 3. 4√(26y3) x 5√(45z5) Answers 1. √64 x √38 64/2 x 38/2 62 x 34 36 x 81 2916 2. √73 x √35 √7271 x √3431 72/2 √7 x 34/2√3 71 √7 x 32√3 (7 x 9) √(7 x 3) 63 √21 3. 4√(26y3) x 5√(45z5) 4√(26y2y1) x 5√(4441z4z1) 4(26/2)(y2/2)√y x 5(44/2)(z4/2)√(4z) 4(8)(y) √y x 5(16)(z2) √(4z) 32y √y x 80z2 √4z (32 x 80)y z2 √( y x 4z) 2,560 yz2 √4yz
|
More TopicsEducational Videos |