Quadratic Inequalities - Examples, Exercises and Solutions

To solve this problem, let's examine the inequality $x^2 + 4 > 0$.

The expression $x^2 + 4$ consists of two terms: $x^2$ and $4$. Notice that:

• The term $x^2$ is always non-negative, which means $x^2 \geq 0$ for any real number $x$.
• The constant term $4$ is positive.

Combining these observations, we see that:

• Since $x^2$ is non-negative, $x^2 + 4 \geq 4$.
• Therefore, $x^2 + 4$ is always greater than zero, as adding 4 to a non-negative number will always yield a positive result.

Thus, there are no values of $x$ for which the expression $x^2 + 4$ is zero or negative. Instead, the expression is always positive for all real numbers $x$.

Therefore, the solution to the inequality $x^2 + 4 > 0$ is all values of $x$.

Let's explore this problem step-by-step:

The inequality given is $x^2 + 9 > 0$.

1. To understand this inequality, we start by considering the expression $x^2$. We know that for any real number $x$, $x^2 \geq 0$. This means $x^2$ is always non-negative.

2. Since $x^2 \geq 0$ for every real number, adding 9 to $x^2$ will necessarily make the expression greater than zero, because a non-negative number plus a positive number gives a positive result: $x^2 + 9 \geq 9 > 0$.

3. Therefore, the inequality $x^2 + 9 > 0$ holds true for all real numbers $x$. There is no value of $x$ that makes the left side equal to or less than zero.

4. Thus, the solution to the inequality is that it holds for all values of $x$.

Consequently, the correct choice from the options provided is:

• All values of $x$

Therefore, the solution is that the inequality $x^2 + 9 > 0$ is true for all values of $x$.

To solve the inequality $-x^2 + 2x > 0$, we begin by considering the corresponding equation $-x^2 + 2x = 0$.

First, factor the quadratic equation:

• Rearrange the terms: $-x^2 + 2x = 0$ becomes $x(2 - x) = 0$.
• This gives us the roots $x = 0$ and $x = 2$.

These roots divide the number line into three intervals: $x < 0$, $0 < x < 2$, and $x > 2$.

We need to test these intervals to determine where the inequality holds:

• For $x < 0$, choose a test point like $x = -1$: the expression $-(-1)^2 + 2(-1) = -1 - 2 = -3$, which is not greater than zero.
• For $0 < x < 2$, choose a test point like $x = 1$: the expression $-(1)^2 + 2(1) = -1 + 2 = 1$, which is greater than zero.
• For $x > 2$, choose a test point like $x = 3$: the expression $-(3)^2 + 2(3) = -9 + 6 = -3$, which is not greater than zero.

Thus, the inequality $-x^2 + 2x > 0$ is satisfied for the interval $0 < x < 2$.

Therefore, the solution to the inequality is $\mathbf{0 < x < 2}$, which corresponds to choice 2 in the given options.

To solve this quadratic inequality, $-x^2 - 9 > 0$, we will follow these steps:

• Step 1: Identify the quadratic expression $-x^2 - 9$.
• Step 2: Attempt transformation and determine when the expression $-x^2 - 9$, can be greater than zero.

Let's analyze the equation:

Rewrite the inequality:
$-x^2 - 9 > 0$

Add 9 to both sides:
$-x^2 > 9$

Multiply the entire inequality by $-1$ and remember to reverse the inequality sign:
$x^2 < -9$

Observe the inequality $x^2 < -9$:
Note that $x^2$, being a square of any real number, is always greater than or equal to zero.

As $x^2$ cannot be less than negative nine for any real number $x$, the inequality has no solution in the realm of real numbers.

Therefore, the correct answer is:

There is no solution.

Let's proceed to solve the inequality $x^2 - 8x + 12 > 0$.

• Start by factoring the quadratic: $x^2 - 8x + 12$.
• Identify the factors of 12 that sum to 8: $6$ and $2$. This results in: $(x - 6)(x - 2) = 0$.

The factorization gives us the critical points $x = 6$ and $x = 2$. These points divide the number line into three intervals: $x < 2$, $2 < x < 6$, and $x > 6$.

Now, we evaluate the sign of the product $(x - 6)(x - 2)$ in each interval:

• For $x < 2$: Both $(x - 6)$ and $(x - 2)$ are negative, so their product is positive.
• For $2 < x < 6$: $(x - 2)$ is positive, $(x - 6)$ is negative, so their product is negative.
• For $x > 6$: Both $(x - 6)$ and $(x - 2)$ are positive, so their product is positive.

The inequality $(x - 6)(x - 2) > 0$ holds for $x < 2$ and $x > 6$.

Thus, the solution to the inequality $x^2 - 8x + 12 > 0$ is $x < 2$ or $x > 6$.

Therefore, the correct answer is $\boxed{x < 2, 6 < x}$.