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 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.

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.

The problem requires us to solve the inequality $x^2 - 3x + 4 < 0$.

To solve the inequality, we first consider the corresponding quadratic equation $x^2 - 3x + 4 = 0$ and find its roots.

Calculate the discriminant $\Delta$:
$\Delta = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7$.

The discriminant $\Delta = -7$ is less than zero, indicating that the quadratic equation has no real roots. This implies that the quadratic expression $x^2 - 3x + 4$ does not change sign and is either always positive or always negative.

Next, evaluate the sign of $x^2 - 3x + 4$. For $x = 0$, the expression is $0^2 - 3 \cdot 0 + 4 = 4$, which is positive. Therefore, the expression is always positive for all real $x$.

Since $x^2 - 3x + 4$ is always positive, there is no $x$ for which $x^2 - 3x + 4 < 0$ holds true.

Therefore, the solution to the inequality is that there is no solution, which corresponds to option 4: "There is no solution."

To solve this inequality $x^2 - 6x + 8 < 0$, we first identify the roots of the equation $x^2 - 6x + 8 = 0$.

Using the quadratic formula, where $a = 1$, $b = -6$, and $c = 8$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

$x = \frac{6 \pm \sqrt{36 - 32}}{2}$

$x = \frac{6 \pm \sqrt{4}}{2}$

$x = \frac{6 \pm 2}{2}$

The solutions are:

$x = \frac{6 + 2}{2} = 4 \quad \text{and} \quad x = \frac{6 - 2}{2} = 2$

The roots are $x = 2$ and $x = 4$. These divide the number line into three intervals: $(-\infty, 2)$, $(2, 4)$, and $(4, \infty)$.

We test each interval to determine where the inequality is satisfied:

• In the interval $(-\infty, 2)$, select $x = 0$. Then:

$0^2 - 6 \times 0 + 8 = 8$, which is greater than 0. Inequality not satisfied.

• In the interval $(2, 4)$, select $x = 3$. Then:

$3^2 - 6 \times 3 + 8 = 9 - 18 + 8 = -1$, which is less than 0. Inequality satisfied.

• In the interval $(4, \infty)$, select $x = 5$. Then:

$5^2 - 6 \times 5 + 8 = 25 - 30 + 8 = 3$, which is greater than 0. Inequality not satisfied.

Therefore, the solution to the inequality $x^2 - 6x + 8 < 0$ is the interval $(2, 4)$.

Thus, the correct answer is $2 < x < 4$.

To solve this quadratic inequality, follow these steps:

• Step 1: Solve the corresponding equation $-x^2 + 3x + 4 = 0$ to find critical points.
• Step 2: Test intervals between critical points to determine where the inequality holds.

Step 1: Solve the equation. The given quadratic is $-x^2 + 3x + 4 = 0$. Let's rewrite it as $x^2 - 3x - 4 = 0$ by multiplying through by $-1$.

We use the quadratic formula where $a = 1$, $b = -3$, and $c = -4$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)}$

$x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}$

The solutions to this equation are:

$x_1 = \frac{3 + 5}{2} = 4$ and $x_2 = \frac{3 - 5}{2} = -1$

Step 2: Determine where the expression is positive by checking intervals:

• Interval $(-\infty, -1)$: Choose $x = -2$. Calculating the expression: $-(-2)^2 + 3(-2) + 4 = -4 - 6 + 4 = -6$ (negative).
• Interval $(-1, 4)$: Choose $x = 0$. Calculating: $-0^2 + 3(0) + 4 = 4$ (positive).
• Interval $(4, \infty)$: Choose $x = 5$. Calculating: $-5^2 + 3(5) + 4 = -25 + 15 + 4 = -6$ (negative).

The quadratic expression $-x^2 + 3x + 4$ is positive in the interval $(-1, 4)$. Hence, for the inequality $-x^2 + 3x + 4 > 0$, we have:

The solution to the inequality is $-1 < x < 4$.

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 $\boxed{x < 2, 6 < x}$.

To solve the inequality $x^2 + 4x > 0$, we will:

• Step A: Find the roots of the equation $x^2 + 4x = 0$.
• Step B: Factor the quadratic to $x(x + 4) = 0$, giving roots $x = 0$ and $x = -4$.
• Step C: Use these roots to break the number line into intervals: $(-\infty, -4)$, $(-4, 0)$, and $(0, \infty)$.
• Step D: Test an arbitrary value from each interval in the inequality $x^2 + 4x > 0$.

Now, let's examine these intervals:

• For $(-\infty, -4)$, choose $x = -5$:
  $(-5)^2 + 4(-5) = 25 - 20 = 5 > 0$. This interval satisfies the inequality.
• For $(-4, 0)$, choose $x = -2$:
  $(-2)^2 + 4(-2) = 4 - 8 = -4 < 0$. This interval does not satisfy the inequality.
• For $(0, \infty)$, choose $x = 1$:
  $1^2 + 4(1) = 1 + 4 = 5 > 0$. This interval satisfies the inequality.

Therefore, the inequality $x^2 + 4x > 0$ holds true for the intervals $(-\infty, -4)$ and $(0, \infty)$.

Therefore, the solution to the inequality is $x < -4, 0 < x$.

To solve the inequality $x^2 - 9 < 0$, we will perform the following steps:

• Step 1: Factor the inequality $x^2 - 9 = (x - 3)(x + 3)$.
• Step 2: Identify the critical values from the factored expression, which occur at $x = 3$ and $x = -3$.
• Step 3: Use these critical points to divide the number line into intervals: $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$.
• Step 4: Test each interval to determine where the inequality holds:
  • For $x = 0$ in the interval $(-3, 3)$, $(0 - 3)(0 + 3) = -9$, which satisfies $< 0$.
  • For $x = -4$ in $(-\infty, -3)$, $(-4 - 3)(-4 + 3) = 7$, which does not satisfy $< 0$.
  • For $x = 4$ in $(3, \infty)$, $(4 - 3)(4 + 3) = 7$, which does not satisfy $< 0$.

Therefore, the inequality $x^2 - 9 < 0$ holds in the interval $-3 < x < 3$. This means any $x$ that falls between these values will satisfy the inequality.

The correct answer is $\mathbf{-3 < x < 3}$.

The objective is to find the values of $x$ such that the inequality $x^2 - 16 > 0$ is satisfied.

Step 1: Factor the inequality expression.

The expression $x^2 - 16$ can be factored using the difference of squares formula:

$x^2 - 16 = (x - 4)(x + 4)$.

Step 2: Determine the critical points.

Set the factors equal to zero to find the critical points:

• $x - 4 = 0$ gives $x = 4$.
• $x + 4 = 0$ gives $x = -4$.

Step 3: Analyze the sign changes on the number line.

We test the intervals defined by the critical points $-4$ and $4$ on a number line: $(-∞, -4)$, $(-4, 4)$, $(4, ∞)$.

Choose a test point from each interval and substitute into the factored expression to check the sign.

• For $x = -5$ (interval $(-∞, -4)$): $(x - 4)(x + 4) = (-5 - 4)(-5 + 4) = (-9)(-1) > 0$.
• For $x = 0$ (interval $(-4, 4)$): $(x - 4)(x + 4) = (0 - 4)(0 + 4) = (-4)(4) < 0$.
• For $x = 5$ (interval $(4, ∞)$): $(x - 4)(x + 4) = (5 - 4)(5 + 4) = (1)(9) > 0$.

Step 4: Extract the solution.

The inequality $x^2 - 16 > 0$ holds true in the intervals where the product is positive, which are $(-∞, -4) \cup (4, ∞)$.

Therefore, the solution to the inequality is $x < -4$ or $x > 4$.

The correct choice is $x < -4, 4 < x$.

To solve the inequality $x^2 - 2x - 8 > 0$, we first need to find the roots of the related equation $x^2 - 2x - 8 = 0$.

Step 1: Factor the quadratic
The quadratic $x^2 - 2x - 8$ can be factored as $(x - 4)(x + 2)$ because:

• The product is $-8$ and the sum is $-2$.
• Expanding $(x - 4)(x + 2)$, we get:
• $(x - 4)(x + 2) = x^2 + 2x - 4x - 8 = x^2 - 2x - 8$.

Step 2: Identify the roots
Set each factor to zero to find the roots:

• $x - 4 = 0 \Rightarrow x = 4$
• $x + 2 = 0 \Rightarrow x = -2$

Step 3: Determine the intervals
The critical points divide the number line into three intervals: $x < -2$, $-2 < x < 4$, and $x > 4$.

Step 4: Test each interval
Choose test points from each interval to check where $(x - 4)(x + 2) > 0$:

• For $x < -2$, take $x = -3$:
  $(-3 - 4)(-3 + 2) = (-7)(-1) = 7 > 0$
• For $-2 < x < 4$, take $x = 0$:
  $(0 - 4)(0 + 2) = (-4)(2) = -8 < 0$
• For $x > 4$, take $x = 5$:
  $(5 - 4)(5 + 2) = (1)(7) = 7 > 0$

Conclusion:
The solution to the inequality $x^2 - 2x - 8 > 0$ is on the intervals $x < -2$ and $x > 4$.

Final Answer:
The correct answer is: Answers (a) and (c)

To solve the inequality $x^2 - 25 < 0$, follow these steps:

• Step 1: Factor the quadratic expression $x^2 - 25$. This is a difference of squares, which factors as $(x - 5)(x + 5)$.
• Step 2: Identify the critical points where the expression equals zero, i.e., where $(x-5)(x+5) = 0$. The solutions are $x = 5$ and $x = -5$.
• Step 3: Determine the intervals defined by these critical points: $(-∞, -5)$, $(-5, 5)$, and $(5, ∞)$.
• Step 4: Test each interval to see where the inequality $(x-5)(x+5) < 0$ holds:
• Choose a test point from $(-∞, -5)$, such as $x = -6$. Then, $(-6-5)(-6+5) = (-11)(-1) > 0$. This interval does not satisfy the inequality.
• Choose a test point from $(-5, 5)$, such as $x = 0$. Then, $(0-5)(0+5) = (-5)(5) = -25 < 0$. This interval satisfies the inequality.
• Choose a test point from $(5, ∞)$, such as $x = 6$. Then, $(6-5)(6+5) = (1)(11) > 0$. This interval does not satisfy the inequality.
• Step 5: Conclude the solution. The inequality holds true in the interval $(-5, 5)$.

Therefore, the solution to the inequality $x^2 - 25 < 0$ is $-5 < x < 5$.

To solve the inequality $-x^2 - 25 < 0$, we start by simplifying it. Rearrange the inequality:

$-x^2 < 25$

Multiplying through by $-1$ (reversing the inequality sign), we have:

$x^2 > -25$

Since $x^2$ is always non-negative for all real numbers (i.e., $x^2 \geq 0$), the smallest value $x^2$ can take is $0$. Therefore, $x^2$ is always greater than $-25$, since any non-negative number is greater than a negative number. Thus, this inequality holds true for all real values of $x$.

Therefore, the solution to the inequality $-x^2 - 25 < 0$ is

All values of $x$.

To solve the inequality $x^2 + 6x > 0$, follow these steps:

• Step 1: Write the inequality in factored form.
  Express $x^2 + 6x$ as $x(x + 6)$.
• Step 2: Identify the roots of the equation $x(x + 6) = 0$.
  The roots are $x = 0$ and $x = -6$.
• Step 3: Determine the sign of $x(x + 6)$ in each interval divided by the roots.
• Step 4: Test three intervals: $x < -6$, $-6 < x < 0$, and $x > 0$.

For $x < -6$:
Pick a value such as $x = -7$. Substituting, $x(x + 6) = (-7)(-7 + 6) = (-7)(-1) = 7 > 0$.
Thus, $x^2 + 6x > 0$ for $x < -6$.

For $-6 < x < 0$:
Pick a value such as $x = -3$. Substituting, $x(x + 6) = (-3)(-3 + 6) = (-3)(3) = -9 < 0$.
Thus, $x^2 + 6x < 0$ for $-6 < x < 0$.

For $x > 0$:
Pick a value such as $x = 1$. Substituting, $x(x + 6) = (1)(1 + 6) = 1 \times 7 = 7 > 0$.
Thus, $x^2 + 6x > 0$ for $x > 0$.

Therefore, the solution to the inequality is $x < -6$ or $x > 0$.

Thus, the correct answer is $x < -6, 0 < x$.