Finding The Range Of K For No Real Roots In Absolute Value Equations

Hey guys! Let's dive into a super interesting problem today. We're going to figure out the range of values for k in the equation

$$-2x + k = 3|x - 4| + 6$$

so that this equation has absolutely no real roots. It's like we're on a quest to find the k that makes this equation impossible to solve in the real world. We'll also explore why a common method – squaring both sides – can sometimes lead us down the wrong path. Buckle up, it's gonna be a fun ride!

The Challenge: No Real Roots

Our main goal here is to determine the range of possible values for the constant k such that the equation $-2x + k = 3|x - 4| + 6$ yields no real solutions for x. This means we need to find the values of k that make the equation have no intersection points when we think about it graphically, or no solutions that satisfy the equation algebraically. Sounds like a puzzle, right? Let's get to solving it!

Isolating the Absolute Value

The first step, which you've already nailed, is to isolate the absolute value part of the equation. This is a classic move in absolute value problems, kind of like setting the stage for the main act. By isolating the absolute value, we can better see what's going on and make strategic decisions. You've correctly started by rearranging the equation like this:

$$-2x + (k - 6) = 3|x - 4|$$

Now, let's take it a step further and divide both sides by 3 to get the absolute value term completely on its own:

$$\frac{-2x + (k - 6)}{3} = |x - 4|$$

This form is super helpful because it sets us up for a graphical approach, which is one of the most intuitive ways to solve this type of problem. We can think of the left side as a line and the right side as a V-shaped graph formed by the absolute value. The solutions to the equation are the x-coordinates of the points where these two graphs intersect. If we want no real roots, we want these graphs to never touch. Think of it as two ships passing in the night – they’re in the same waters, but they never collide.

Graphical Interpretation

Visualizing the Problem: Graphically, the absolute value equation $|x - 4|$ represents a V-shaped graph with its vertex at the point (4, 0). The equation $\frac{-2x + (k - 6)}{3}$ represents a straight line with a slope of $-\frac{2}{3}$ and a y-intercept that depends on the value of k. The solutions to the original equation are the x-coordinates of the points where the line intersects the V-shaped graph. For the equation to have no real roots, the line and the V-shaped graph must not intersect at any point. Think of it like this: the line needs to either be completely above the V or completely below it, without ever touching it.

Analyzing the V-Shape: The V-shaped graph of $|x - 4|$ has two parts: one where $x < 4$ and another where $x \geq 4$. For $x < 4$, the absolute value $|x - 4|$ is equivalent to $-(x - 4)$, and for $x \geq 4$, it is equivalent to $(x - 4)$. This means our V-shape is made up of two lines joined at the vertex (4, 0). The line on the left has a slope of -1, and the line on the right has a slope of +1.

The Role of the Line: Now, let's consider the line $\frac{-2x + (k - 6)}{3}$. As we mentioned, this is a straight line with a slope of $-\frac{2}{3}$. The key here is the y-intercept, which changes as k changes. We need to find the range of k values that make this line miss the V-shape entirely. Imagine sliding this line up and down – sometimes it will cross the V, and sometimes it won't. We want to find the sweet spot where it doesn't.

Finding the Critical Points: To do this, we need to focus on the critical points where the line would just barely touch the V-shape. These points occur where the line is tangent to one of the sides of the V. Since our line has a negative slope, it will intersect the left side of the V-shape (where $x < 4$) if it intersects at all. The critical point is when the line touches the V exactly at its corner, the vertex (4, 0). If the line is above this point, it won't intersect; if it's below, it will.

Setting Up the Condition: So, we need to find the value of k that makes the line pass through the point (4, 0). Plugging this point into our line equation gives us:

$$\frac{-2(4) + (k - 6)}{3} = 0$$

Simplifying this, we get:

$$-8 + k - 6 = 0$$

$$k = 14$$

This is a crucial value. When k = 14, the line passes right through the vertex of the V. For the line to have no intersection, it needs to be above this point. This means that k must be greater than 14.

The Final Range: Therefore, the range of possible values for k such that the equation has no real roots is k > 14. This is the answer we've been searching for – the range of k values that makes our equation unsolvable in the real number system. High five!

The Pitfall of Squaring Both Sides

Now, let's tackle the second part of the question: Why does squaring both sides sometimes lead to an incorrect answer? This is a common mistake, and understanding why it happens is super important for mastering absolute value equations. Squaring both sides is a tempting move because it eliminates the absolute value, which can feel like a big win. But, like any shortcut, it has its risks.

The Temptation of Squaring

When faced with an equation like

$$-2x + k = 3|x - 4| + 6$$

you might think, “Aha! Let's just square both sides and get rid of that pesky absolute value!” This would give you:

$$(-2x + k - 6)^2 = (3|x - 4|)^2$$

$$(-2x + k - 6)^2 = 9(x - 4)^2$$

This looks cleaner, right? No more absolute value! But here's the catch: squaring both sides can introduce extraneous solutions. Extraneous solutions are solutions that you get algebraically, but they don't actually satisfy the original equation. They're like imposters – they sneak into your solution set, but they're not the real deal.

Why Extraneous Solutions Appear

The reason extraneous solutions pop up is because squaring a number makes it positive. In other words, both a positive number and its negative will have the same square. For example, both 3 and -3, when squared, become 9. This means that when you square both sides of an equation, you're essentially saying, “If A = B, then A² = B²,” which is true. However, the reverse isn't necessarily true. If A² = B², it doesn't automatically mean that A = B; it could also mean that A = -B. Think of it like this: if you know two numbers have the same square, you don't know if they started out as the same number or as opposites.

Applying this to our Problem: When we square both sides of our equation, we're essentially treating the cases where $-2x + k - 6 = 3(x - 4)$ and $-2x + k - 6 = -3(x - 4)$ as the same. The absolute value neatly handles both of these cases, but squaring blurs the distinction. This is why squaring can lead to solutions that work in the squared equation but not in the original equation with the absolute value.

A Concrete Example

To make this clearer, let’s consider a simpler example:

$$x = |x - 1|$$

If we square both sides, we get:

$$x^2 = (x - 1)^2$$

$$x^2 = x^2 - 2x + 1$$

$$2x = 1$$

$$x = \frac{1}{2}$$

So, we get $x = \frac{1}{2}$ as a potential solution. But let's plug it back into the original equation:

$$\frac{1}{2} = |\frac{1}{2} - 1|$$

$$\frac{1}{2} = |-\frac{1}{2}|$$

$$\frac{1}{2} = \frac{1}{2}$$

This solution works! However, let's change the original equation slightly to illustrate an extraneous solution:

$$x = -|x - 1|$$

If we square both sides, we get the exact same squared equation as before:

$$x^2 = (x - 1)^2$$

And we still get $x = \frac{1}{2}$ as a potential solution. But now, let's plug it back into this new original equation:

$$\frac{1}{2} = -|\frac{1}{2} - 1|$$

$$\frac{1}{2} = -|-\frac{1}{2}|$$

$$\frac{1}{2} = -\frac{1}{2}$$

This is clearly false! So, $x = \frac{1}{2}$ is an extraneous solution in this case. It snuck in because squaring both sides lost the crucial negative sign.

The Takeaway

The key takeaway here is that squaring both sides can be a valid technique, but you must check your solutions in the original equation. This is like being a detective and double-checking your evidence to make sure it all adds up. If a solution doesn't work in the original equation, it's an extraneous solution, and you need to discard it. In our original problem, squaring both sides would lead to a more complicated algebraic mess, and it would be easy to miss the extraneous solutions, making the graphical approach much more reliable and clear.

Wrapping Up

So, guys, we've successfully navigated the challenge of finding the range of k for our absolute value equation to have no real roots. We discovered that k must be greater than 14. We also explored why squaring both sides can lead to extraneous solutions, and why it's a method to approach with caution. Remember, understanding the underlying principles and visualizing the problem graphically can often save you from algebraic pitfalls. Keep up the awesome work, and happy problem-solving!