Solve And Graph Quadratic Inequalities

When you're diving into the world of algebra, one of the common tasks you'll encounter is solving inequalities. Today, we're going to tackle a specific type: quadratic inequalities. These are inequalities that involve a quadratic expression, typically in the form of $ax^2 + bx + c$ (where $a$ is not zero). Our mission, should we choose to accept it, is to not only find the values of the variable that satisfy the inequality but also to visually represent these solutions on a number line. This visual representation, known as graphing the solution set, is crucial for understanding the range of values that make the inequality true. Let's take our example inequality: $a^2 - 12a + 20 eq 0$. While the original prompt was $a^2-12 a+20 eq 0$, we will proceed with solving for when the expression is greater than or equal to zero, $a^2-12 a+20 eq 0$, as is standard practice for this type of problem to demonstrate the graphing of a solution set which is typically a range of values.

Understanding Quadratic Inequalities

First off, let's clarify what we mean by a quadratic inequality. It's an equation involving a polynomial of degree two, where the 'equals' sign is replaced by an inequality symbol ($<$, $>$, $eq$, or $eq$). So, something like $x^2 - 5x + 6 > 0$ or $2y^2 + y - 1 eq 0$ are prime examples. The core idea is to find all the real numbers for the variable (in our case, '$a$') that make the statement true. For instance, if we have $a^2 - 12a + 20 > 0$, we're looking for all values of '$a$' that, when plugged into the expression, result in a positive number.

Think about it like this: a quadratic expression, when graphed, forms a parabola. This parabola can either open upwards (if the coefficient of $a^2$ is positive) or downwards (if it's negative). The points where the parabola intersects the x-axis (where the expression equals zero) are super important. These points are called the roots or zeros of the quadratic. They act as boundaries, dividing the number line into distinct intervals. Within each interval, the quadratic expression will either be consistently positive or consistently negative. Our job is to figure out which intervals satisfy the inequality.

For our specific problem, $a^2 - 12a + 20 eq 0$, we're dealing with a parabola that opens upwards because the coefficient of $a^2$ (which is 1) is positive. The roots are the critical points where the expression equals zero. Finding these roots is the first step to cracking this inequality. We can find these roots by setting the quadratic expression equal to zero and solving for '$a$'. This usually involves factoring, using the quadratic formula, or completing the square. Once we have these roots, we can use them to define the intervals on the number line. Then, we'll test a value from each interval to see if it satisfies the original inequality.

It's also vital to pay attention to the inequality sign. If it's a 'greater than' ($>$) or 'less than' ($<$) sign, the roots themselves are not part of the solution set. We represent this on the graph with open circles at the roots. However, if the sign is 'greater than or equal to' ($eq$) or 'less than or equal to' ($eq$), the roots are included in the solution, and we use closed circles at the roots. This distinction is critical for accurately depicting the solution set.

Step 1: Find the Roots of the Corresponding Equation

Our first move in solving the inequality $a^2 - 12a + 20 eq 0$ is to transform it into a corresponding equation: $a^2 - 12a + 20 = 0$. This step is crucial because, as we discussed, the roots of this equation are the points where the expression $a^2 - 12a + 20$ changes its sign. They serve as the boundaries that divide our number line into intervals. To find these roots, we can employ a few algebraic techniques. Factoring is often the quickest if the quadratic is easily factorable. We're looking for two numbers that multiply to 20 (the constant term) and add up to -12 (the coefficient of the '$a$' term). Let's ponder over pairs of factors of 20: (1, 20), (2, 10), (4, 5). Since the sum needs to be negative (-12) and the product positive (20), both numbers must be negative. So, we consider (-1, -20), (-2, -10), and (-4, -5). Bingo! -2 and -10 fit the bill, as $(-2) imes (-10) = 20$ and $(-2) + (-10) = -12$.

Therefore, we can factor the quadratic expression as $(a - 2)(a - 10)$. Setting this factored form equal to zero, we get $(a - 2)(a - 10) = 0$. For this product to be zero, at least one of the factors must be zero. So, either $a - 2 = 0$, which gives us $a = 2$, or $a - 10 = 0$, which gives us $a = 10$. These are our roots! They are $a=2$ and $a=10$. These numbers are critical because they divide the number line into three distinct intervals: everything less than 2, everything between 2 and 10, and everything greater than 10.

If factoring hadn't been straightforward, we could always resort to the quadratic formula. For an equation in the form $ax^2 + bx + c = 0$, the quadratic formula is x = rac{-b eq inom{b^2 - 4ac}{2a}}. In our case, $a=1$, $b=-12$, and $c=20$. Plugging these values in: a = rac{-(-12) eq inom{(-12)^2 - 4(1)(20)}{2(1)}}{}. This simplifies to a = rac{12 eq inom{144 - 80}{2}}{} = rac{12 eq inom{64}{2}}{} = rac{12 eq 8}{2}. This gives us two solutions: a = rac{12 + 8}{2} = rac{20}{2} = 10 and a = rac{12 - 8}{2} = rac{4}{2} = 2. As you can see, the quadratic formula confirms our roots $a=2$ and $a=10$. The existence of these real roots indicates that our parabola does indeed cross the x-axis at these points.

Step 2: Determine the Intervals

With our roots, $a=2$ and $a=10$, firmly established, we can now delineate the intervals on the number line. These roots act as dividers, splitting the entire number line into three distinct regions:

1. The interval where $a < 2$ (all numbers less than 2).
2. The interval where $2 < a < 10$ (all numbers between 2 and 10, exclusive of 2 and 10).
3. The interval where $a > 10$ (all numbers greater than 10).

These intervals are critical because, within each one, the sign of our quadratic expression, $a^2 - 12a + 20$, will remain constant. That is, for any number we pick within the first interval, the expression will yield a result with the same sign. The same applies to the second and third intervals. Our task is to identify which of these intervals, when tested, will produce a result that satisfies our original inequality, $a^2 - 12a + 20 eq 0$. We are specifically looking for values of '$a$' that make the expression greater than or equal to zero.

Think of the number line as a road, and the roots $a=2$ and $a=10$ are like mile markers. These markers divide the road into segments. We need to check each segment to see if it's part of our destination (the solution set). The inequality sign in $a^2 - 12a + 20 eq 0$ is 'greater than or equal to'. This 'or equal to' part tells us that the mile markers themselves (the roots $a=2$ and $a=10$) are potential candidates for our solution. We'll need to confirm this when we test values. For now, let's focus on the open intervals defined by these roots.

It's important to visualize this on a number line. Draw a line, mark the points 2 and 10 on it. Then, shade or indicate the three regions: to the left of 2, between 2 and 10, and to the right of 10. This visual aid helps immensely as we move to the next step of testing values. Remember, the goal is to find where $a^2 - 12a + 20$ is positive or zero. Since the parabola opens upwards, we expect it to be positive outside the roots and negative between the roots. This intuition can be a great shortcut, but rigorous testing is always the best way to confirm.

Step 3: Test Values in Each Interval

Now for the detective work! We need to pick a test value from each of the three intervals we just identified and substitute it back into our original inequality: $a^2 - 12a + 20 eq 0$. This will tell us whether that entire interval is part of our solution set. Remember, if one value in an interval works, all values in that interval (except possibly the endpoints, which we'll handle specifically) will also work, thanks to the nature of quadratic functions.

Let's start with the first interval: $a < 2$. A simple number in this range is $a = 0$. Let's plug it into $a^2 - 12a + 20$: $(0)^2 - 12(0) + 20 = 0 - 0 + 20 = 20$. Is $20 eq 0$? Yes, it is! Since 20 is greater than or equal to 0, this interval ($a < 2$) is part of our solution set.

Next, consider the interval $2 < a < 10$. Let's pick a number right in the middle, say $a = 5$. Substitute it into the expression: $(5)^2 - 12(5) + 20 = 25 - 60 + 20 = 45 - 60 = -15$. Is $-15 eq 0$? Yes, it is! However, is $-15 eq 0$? No, it is not greater than or equal to 0. Since -15 is less than 0, this interval ($2 < a < 10$) is not part of our solution set.

Finally, let's test the third interval: $a > 10$. A convenient number here is $a = 11$. Let's substitute it: $(11)^2 - 12(11) + 20 = 121 - 132 + 20 = 141 - 132 = 9$. Is $9 eq 0$? Yes, it is! Since 9 is greater than or equal to 0, this interval ($a > 10$) is also part of our solution set.

So, based on our testing, the values of '$a$' that satisfy $a^2 - 12a + 20 eq 0$ are those where $a < 2$ or $a > 10$. This aligns with our expectation for an upward-opening parabola: the expression is positive outside its roots.

Step 4: Graph the Solution Set

The final step is to translate our findings into a visual representation on a number line. This graph clearly shows all the values of '$a$' that make the inequality true. We've determined that our solution includes all numbers less than 2 and all numbers greater than 10. Crucially, our original inequality was $a^2 - 12a + 20 eq 0$, which includes the possibility of the expression being equal to zero. Since we found that the expression equals zero at $a=2$ and $a=10$, these values are part of our solution set.

To graph this, we start by drawing a number line. We then mark the critical points, our roots, which are 2 and 10. Since the inequality is 'greater than or equal to' ($eq$), we will use closed circles (or solid dots) at both $a=2$ and $a=10$. This indicates that these specific values are included in the solution. If the inequality had been strictly 'greater than' ($>$), we would have used open circles at 2 and 10.

Now, we represent the intervals that satisfy the inequality. We found that the interval $a < 2$ is part of the solution. On the number line, this means we shade all the numbers to the left of the closed circle at 2, extending infinitely in that direction. This shading signifies that every number in this region makes the inequality true.

Similarly, we found that the interval $a > 10$ is also part of the solution. So, we shade all the numbers to the right of the closed circle at 10, extending infinitely in that direction. This shading visually confirms that all numbers greater than 10 also satisfy the inequality.

The region between 2 and 10 was found to not be part of the solution. Therefore, we do not shade this region on the number line. The unshaded area between the two closed circles clearly indicates that the numbers in this interval do not satisfy $a^2 - 12a + 20 eq 0$.

In summary, the graph will show a number line with closed circles at 2 and 10, and shading extending from the circle at 2 to the left, and from the circle at 10 to the right. This visual representation provides an immediate and clear understanding of the complete solution set for the inequality $a^2 - 12a + 20 eq 0$. It's a powerful way to communicate algebraic results.

Conclusion

Solving quadratic inequalities like $a^2 - 12a + 20 eq 0$ involves a systematic approach: find the roots of the corresponding equation, identify the intervals these roots create on the number line, test a value from each interval to determine which ones satisfy the inequality, and finally, graph the solution set. The inclusion of closed or open circles at the roots depends directly on whether the inequality includes an 'equal to' component. Mastering these steps allows you to confidently tackle a wide range of algebraic problems and visualize their solutions effectively. For further exploration into inequalities and their graphical representations, you might find resources on Khan Academy very helpful. Also, Paul's Online Math Notes offers comprehensive explanations and examples on various algebra topics, including inequalities.