We need to determine whether each of the given statements is true or false. i) The number $\pi^0$ is irrational. ii) The quadratic equation $x^2 = -4$ has no real solutions. iii) The rational expression $\frac{3x-1}{1-3x}$ when simplified is equal to $-1$. iv) $x^{5/3} = \sqrt[5]{x^3}$ for a real number $x$. v) The ordered pair $(-2, 5)$ is a solution of the system $y - x = 7$ $2y + 5x = 0$ vi) To graph the inequality $3x < -2y$ the ordered pair $(0, 0)$ should be used as a test point. vii) The polynomial $x^2 + 4$ factors as $(x + 2)^2$. viii) $\sqrt[n]{x}$ is never negative when $n$ is an even natural number.

1. Problem Description

We need to determine whether each of the given statements is true or false.

i) The number $\pi^0$ is irrational.

ii) The quadratic equation $x^2 = -4$ has no real solutions.

iii) The rational expression $\frac{3x-1}{1-3x}$ when simplified is equal to $-1$.

iv) $x^{5/3} = \sqrt[5]{x^3}$ for a real number $x$.

v) The ordered pair $(-2, 5)$ is a solution of the system

$y - x = 7$

$2y + 5x = 0$

vi) To graph the inequality $3x < -2y$ the ordered pair $(0, 0)$ should be used as a test point.

vii) The polynomial $x^2 + 4$ factors as $(x + 2)^2$.

viii) $\sqrt[n]{x}$ is never negative when $n$ is an even natural number.

2. Solution Steps

i) Any number raised to the power of 0 is

1. So, $\pi^0 = 1$. 1 is a rational number. Therefore, the statement is false.

ii) The equation is $x^2 = -4$. Taking the square root of both sides, we get $x = \pm\sqrt{-4} = \pm 2i$. These are imaginary solutions. Thus, there are no real solutions. The statement is true.

iii) The rational expression is $\frac{3x-1}{1-3x}$. We can rewrite the denominator as $-(3x-1)$. Then the expression becomes $\frac{3x-1}{-(3x-1)}$. If $3x-1 \ne 0$, which means $x \ne 1/3$, then $\frac{3x-1}{-(3x-1)} = -1$. Therefore, the statement is true.

iv) $x^{5/3}$ can be written as $\sqrt[3]{x^5}$, not $\sqrt[5]{x^3}$. Thus the statement is false.

$x^{5/3} = \sqrt[3]{x^5} \neq \sqrt[5]{x^3}$

v) The system of equations is

$y - x = 7$

$2y + 5x = 0$

Substitute $x = -2$ and $y = 5$ into the equations:

$5 - (-2) = 5 + 2 = 7$, which is true.

$2(5) + 5(-2) = 10 - 10 = 0$, which is true.

Therefore, $(-2, 5)$ is a solution of the system. The statement is true.

vi) To determine if the ordered pair $(0, 0)$ should be used as a test point, we substitute it into the inequality $3x < -2y$.

$3(0) < -2(0)$ which simplifies to $0 < 0$, which is false. Since the inequality is strict ($<$), the line $3x = -2y$ is not included in the solution. Since $(0,0)$ lies on the line $3x=-2y$, it should not be used as a test point. If we choose a point on the line $3x=-2y$, the origin $(0, 0)$, it will not work because we cannot decide which side of the line is the correct solution to the inequality. Therefore, the statement is true.

vii) $(x + 2)^2 = x^2 + 4x + 4$. This is not equal to $x^2 + 4$. Thus, the statement is false.

viii) If $n$ is an even natural number, then $\sqrt[n]{x}$ is only defined for non-negative values of $x$. If $x$ is positive, then $\sqrt[n]{x}$ will also be positive. If $x=0$, then $\sqrt[n]{0} = 0$. Therefore, $\sqrt[n]{x}$ is never negative when $n$ is an even natural number. The statement is true.

3. Final Answer

i) F

ii) T

iii) T

iv) F

v) T

vi) T

vii) F

viii) T