Write each of the following in the form a √ b or a ⁿ√ b, where a and b are integers and b has the least value possible:
a. √242 b. √288 c. √360 d. ⁿ√162
a) ¡î(2*121) = ¡î121 ¡î2 = 11¡î2
b) ¡î(144*2) = ¡î144 ¡î2 = 12¡î2
c) ¡î(36*10) = 6¡î10
d) not so sure about d...
is ©ú¡î a typo?
The font didn't seem to process, so i did it in txt form again:
a) sqrt(2*121) = sqrt(121) sqrt(2) = 11*sqrt(2)
b) sqrt(144*2) = sqrt(144) sqrt(2) = 12*sqrt(2)
c) sqrt(36*10) = 6*sqrt(10)
d) not so sure about d...
is nroot(162) a typo?
a. To write √242 in the form a √ b, we need to determine the smallest perfect square (b) that can be factored out from 242. The prime factorization of 242 is 2 * 11 * 11. To get the smallest value of b, we take the square root of 2 * 11 * 11. The square root of 2 is (√2) and the square root of 11 * 11 is 11. Therefore, we can write √242 as √(2 * 11 * 11) = 11√2.
b. To write √288 in the form a √ b, we find the prime factorization of 288: 2 * 2 * 2 * 2 * 3 * 3. We group the factors into pairs of equal numbers. In this case, we have (2 * 2), (2 * 2), and (3 * 3). We take the square root of each group. The square root of (2 * 2) is 2 and the square root of (3 * 3) is 3. Therefore, we can write √288 as √(2 * 2 * 2 * 2 * 3 * 3) = 2 * 2 * 3 = 12√2.
c. To write √360 in the form a √ b, we find the prime factorization of 360: 2 * 2 * 2 * 3 * 3 * 5. Similar to the previous example, we group the factors into pairs of equal numbers. We have (2 * 2), (3 * 3), and 5. The square root of (2 * 2) is 2 and the square root of (3 * 3) is 3. Therefore, we can write √360 as √(2 * 2 * 2 * 3 * 3 * 5) = 2 * 3 * √5 = 6√5.
d. To write ⁿ√162 in the form a ⁿ√ b, we need to determine the smallest value of b. Since the question does not specify the value of n, we assume n = 2. The prime factorization of 162 is 2 * 3 * 3 * 3 * 3. We group the factors into pairs of equal numbers. We have (2), (3 * 3), and (3 * 3). The ⁿ√(2) is 2 and the ⁿ√(3 * 3) is 3. Therefore, we can write ⁿ√162 as 2 * 3 * 3 = 18.