Mrs. Cathy's Mathematics Class

Powers and Roots

On this page we hope to clear up problems you might have with powers (exponents) and roots (square, cube, n-th). Square roots and exponents are used everywhere in math and can get quite complicated.
Solution: 5 That is the answer because 5 * 5 = 25. 2. Problem:
-SQRT(64) Solution: -8 Note that the negative sign is out- side the square root "symbol."
3. Problem:
SQRT((-16) ^{2}) By the theorem above, the answer is Solution: |-16|. The absolute value of
-16 is .16
The game is in the first site on the list. Remember the highest amount of money you won, in the game, for your quiz.
1. Problem: CBRT(27y
^{3}) Solution: 3y
## The radical is the symbol that represents a square root. -The radicand is the number underneath the radical symbol.
1. Problem: Find the 4th root of 16.
The problem asks us to find a number that raised to the 4thSolution: power equals 16. Write 16 as a product of prime factors. . 2 * 2 * 2 * 2 To be able to remove something from under the radical, there have to be 4 instances of it (because we're taking a 4th root). There are four instances of 2. That leaves nothing under the radical, so the answer is .22. Problem: Find the 10th root of 50. ^{10} By the Solution: , the 10th root of n-th Root Theorem50 is ^{10} |50|, or . 50
1. Multiply:
SQRT(x + 2) * SQRT(x - 2) Solution: Use the theorem above to put both terms under the same radical. SQRT((x + 2)(x - 2)) Multiply the binomials under the radical out. SQRT(x ^{2} - 4) 2. Problem: Simplify
SQRT(20) Solution: Factor the radicand as a product of prime factors. SQRT(2 * 2 * 5)There are two instances of 2, so by the definition of a square root, you can take 2 out from under the radicand. That gives the following for an answer: 2(SQRT(5))
SQRT(27/y, don't be scared of the fraction. Just use the ^{2})Roots of Fractions theorem, which says that nRT(a/b) = (nRT(a))/(nRT(b)). Examples: 1. Simplify:
CBRT(27/125) Solution: Use the Roots of Fractions theorem to rewrite the problem. CBRT(27) --------- CBRT(125) Take the cube root of both the numerator and denominator to get the final answer. (3/5) 2. Simplify:
SQRT(25/y ^{2})Solution: Use the Roots of Fractions theorem to rewrite the problem. SQRT(25) -------- SQRT(y ^{2})Take the square root of both the numerator and denominator to get the final answer. (5/y) 3. Simplify:
(SQRT(80))/(SQRT(5)) Solution: Use the converse of the Roots of Fractions theorem and rewrite it as one radical. SQRT(80/5) simplifies to 80/5 16. SQRT(16) = 4
1. Problem:
6(SQRT(7)) + 4(SQRT(7)) Solution: Both terms are alike (like radicals have the same index and same radicand), so you can add them. 10(SQRT(7)) 2. Problem:
3(SQRT(8)) - 5(SQRT(2)) Solution: Factor 8. 3 (SQRT(4 * 2)) - 5(SQRT(2))Factor SQRT(4 * 2) into two radicals. 3 (SQRT(4))(SQRT(2)) - 5(SQRT(2))Take the square root of 4. 3 * 2(SQRT(2)) - 5(SQRT(2))(SQRT(2)) - 5(SQRT(2))6 Combine like terms. SQRT(2) 3. Problem:
CBRTy * (CBRT(y ^{2}) + CBRT(2))Solution: Use the distributive law of multiplication, which says thata(b + c) = ab + ac to multiply the expression out. CBRTy * CBRT(y ^{2}) + CBRTy * CBRT(2)Multiply the radicals. CBRT (y + CBRT^{3})(2y)Take the cube root of y. ^{3}y + CBRT(2y)
1. Problem: Rationalize the denominator.
4 + SQRT(2) ----------- 5 - SQRT(2) Solution: Multiply by 1 (make sure the fraction you choose to use as one will make the denominator a perfect square — the conjugate is usually a good number). 4 + SQRT(2) 5 + SQRT(2)------------- * ----------- 5 - SQRT(2) 5 + SQRT(2) Multiply the problem as you would multiply any fractions. Also, the FOIL method of multiplying binomials will come in handy. 20 + 4(SQRT(2)) + 5(SQRT(2)) + (SQRT(2)) ^{2}----------------------------------------- 25 + 5(SQRT(2)) - 5(SQRT(2)) - (SQRT(2)). Perform any indicated operations. 20 + 4(SQRT(2)) + 5(SQRT(2)) + 2 -------------------------------- 25 + 5(SQRT(2)) - 5(SQRT(2)) - 2 Perform any indicated operations, and combine like terms, if you can. 22 + 9(SQRT(2)) --------------- 23
Those examples show us that fractions can be used for exponents. However, this could become confusing when we see a number raised to the four-fifths power. Actually, it's not that bad. There is a definition that states, for any natural numbers, m and k, and any nonnegative number a, a^{m/k} equals the k-th root of a^{m}. Example: 1. Problem: Simplify
(27) ^{2/3}Solution: Rewrite using the definition stated above. CBRT(27 ^{2})CBRT(729) 9 2. Problem: Simplify the
6th root of x ^{3}Solution: Rewrite by reversing the definition. x ^{3/6}Use arithmetic to simplify the exponent. x ^{1/2}By definition, x is the same as ^{1/2}. SQRT(x) 3. Problem: Simplify
4 ^{-(1/2)}Solution: Use the definition of negative exponents and rewrite the expression with positive exponents for a simplified expression.
1 ----- Since 4 ^{(1/2)}4 is the same as ^{1/2}SQRT(4) and the square root of 4 is 2, the answer is . 1/2
1. Problem:
SQRT(x) - 3 = 4 Solution: Add 3 to each side. SQRT(x) = 7 Using the Principle of Powers theorem, square both sides.x = 7 ^{2}x = 49 |