1/14 (w#3 find the number of roots
1) $sqrt3{8}$ (4) $sqrt4{256}$
2) $sqrt2{0}$ (5) $sqrt6{-729}$
3) $sqrt5{-1}$

Question

1/14 (w#3 find the number of roots
1) $sqrt3{8}$  (4) $sqrt4{256}$
2) $sqrt2{0}$  (5) $sqrt6{-729}$
3) $sqrt5{-1}$

Accepted Answer

### 1. For $\boldsymbol{\sqrt[3]{8}}$
# Explanation:
## Step1: Recall the definition of nth roots
For a real number $ a $ and positive integer $ n $, the $ n $-th root of $ a $, $ \sqrt[n]{a} $, has a certain number of real roots depending on the parity of $ n $ and the sign of $ a $. For odd $ n $, every real number has exactly one real $ n $-th root. For even $ n $, non - negative real numbers have two real $ n $-th roots (positive and negative) and negative real numbers have no real $ n $-th roots. Also, $ \sqrt[n]{0}=0 $ for any $ n $.
Here, $ n = 3 $ (odd) and $ a=8 $.
## Step2: Determine the number of roots
Since $ n = 3 $ is odd and $ 8$ is a real number, the cube root of $ 8 $ (i.e., $ \sqrt[3]{8} $) has exactly 1 real root (because for odd $ n $, the function $ y = \sqrt[n]{x}$ is a one - to - one function over the real numbers). We know that $ \sqrt[3]{8}=2 $ (since $ 2^3 = 8 $), and there is only one real number that when cubed gives 8.
# Answer: 1

### 2. For $\boldsymbol{\sqrt[2]{0}}$ (which is $\boldsymbol{\sqrt{0}}$)
# Explanation:
## Step1: Recall the property of square root of 0
The square root of a number $ x $ is a number $ y $ such that $ y^2=x $. For $ x = 0 $, we have $ y^2=0 $.
## Step2: Solve for y
If $ y^2 = 0 $, then $ y = 0 $ (since $ 0\times0=0 $ and there is only one number that satisfies this equation for the square root of 0. Although for non - zero non - negative $ x $, $ \sqrt{x} $ has two roots ($ \sqrt{x} $ and $ -\sqrt{x} $), for $ x = 0 $, both the positive and negative roots are 0. So, the square root of 0 has 1 real root.
# Answer: 1

### 3. For $\boldsymbol{\sqrt[5]{-1}}$
# Explanation:
## Step1: Recall the rule for odd nth roots
For an odd positive integer $ n $, every real number has exactly one real $ n $-th root. Here, $ n = 5 $ (which is odd) and $ a=-1 $.
## Step2: Determine the number of roots
Since $ n = 5 $ is odd and $ - 1$ is a real number, the fifth root of $ - 1 $ (i.e., $ \sqrt[5]{-1}$) has exactly 1 real root. We know that $ (-1)^5=-1 $, and there is only one real number that when raised to the 5th power gives $ - 1 $.
# Answer: 1

### 4. For $\boldsymbol{\sqrt[4]{256}}$
# Explanation:
## Step1: Recall the rule for even nth roots of positive numbers
For $ n = 4 $ (even) and $ a = 256>0 $. The $ n $-th root of a positive number $ a $ (where $ n $ is even) has two real roots: $ \sqrt[n]{a}$ and $ -\sqrt[n]{a} $ (except when $ a = 0 $).
First, we find the positive 4th root of 256. We know that $ 4^4=256 $ (since $ 4\times4\times4\times4 = 256 $) and $ (- 4)^4=(-4)\times(-4)\times(-4)\times(-4)=256 $.
## Step2: Count the number of roots
Since $ n = 4 $ is even and $ 256>0 $, the 4th root of 256 has 2 real roots (4 and - 4).
# Answer: 2

### 5. For $\boldsymbol{\sqrt[6]{-729}}$
# Explanation:
## Step1: Recall the rule for even nth roots of negative numbers
For an even positive integer $ n $, the $ n $-th root of a negative number does not exist in the set of real numbers. Here, $ n = 6 $ (even) and $ a=-729<0 $.
## Step2: Determine the number of real roots
Since $ n = 6 $ is even and $ - 729$ is negative, there are no real numbers $ x $ such that $ x^6=-729 $ (because any real number raised to an even power is non - negative). So, the number of real roots of $ \sqrt[6]{-729} $ is 0.
# Answer: 0

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QUESTION IMAGE

Question

Explanation:

1. For $\boldsymbol{\sqrt[3]{8}}$

Step1: Recall the definition of nth roots

Step2: Determine the number of roots

Step1: Recall the property of square root of 0

Step2: Solve for y

Step1: Recall the rule for odd nth roots

Step2: Determine the number of roots

Answer:

2. For $\boldsymbol{\sqrt[2]{0}}$ (which is $\boldsymbol{\sqrt{0}}$)