Approximate Solutions To Equations: A Step-by-Step Guide

Hey guys! Let's dive into the fascinating world of finding solutions to polynomial equations. Sometimes, these solutions aren't neat, whole numbers, and we need to get our hands dirty with approximations. Today, we're tackling a cubic equation, which means it has a term with $x^3$ in it. These can be tricky, but we'll break it down step by step.

The Challenge: A Cubic Equation

Our mission, should we choose to accept it, is to approximate a solution (also called a root) to the equation:

$$4x^3 + 2x^2 - 6x + 2 = 0$$

We're told that there's a solution, which we're calling 'r', hiding somewhere in the interval between -2 and -1. This is super helpful because it gives us a starting point for our search. We need to find 'r' accurate to two decimal places. That means we need to keep going until we're sure the first two digits after the decimal point are correct.

Why do we need approximations? You might be wondering why we can't just solve it directly. Well, there are formulas for solving cubic equations, but they can be quite complex. For many real-world problems, an approximate solution that's accurate enough is perfectly fine, and often easier to obtain. Plus, numerical methods (the techniques we use for approximation) are used extensively in computer programs and scientific calculations.

The Strategy: Intermediate Value Theorem to the Rescue

Our main weapon in this quest is the Intermediate Value Theorem (IVT). This theorem is a cornerstone of calculus and a powerful tool for root-finding. In simple terms, the IVT says: If you have a continuous function (like our polynomial) and you plug in two different values, say 'a' and 'b', and the function gives you results with opposite signs (one positive, one negative), then there must be a root somewhere between 'a' and 'b'.

Think of it like this: Imagine a roller coaster track going up and down. If the track is continuous (no breaks!), and at one point you're below ground (negative) and at another point you're above ground (positive), you had to cross ground level (zero) somewhere in between. That ground level crossing is our root!

Let's call our polynomial function $f(x) = 4x^3 + 2x^2 - 6x + 2$. We know a root exists between -2 and -1. So, let's see what the IVT can tell us. We evaluate our function at the endpoints of the interval:

• $f(-2) = 4(-2)^3 + 2(-2)^2 - 6(-2) + 2 = -32 + 8 + 12 + 2 = -10$
• $f(-1) = 4(-1)^3 + 2(-1)^2 - 6(-1) + 2 = -4 + 2 + 6 + 2 = 6$

As expected, $f(-2)$ is negative and $f(-1)$ is positive. This confirms that there's indeed a root between -2 and -1.

The Hunt: Bisection Method in Action

Now that we know where to look, let's zoom in on that root. We'll use the bisection method, a classic algorithm for finding roots. It's based on the IVT and works by repeatedly halving the interval and choosing the subinterval where the sign change occurs. Basically, we're playing a guessing game, but a smart one!

Here's how it works:

1. Find the midpoint: Calculate the midpoint of our interval. In our first step, the interval is [-2, -1], so the midpoint is $(-2 + (-1))/2 = -1.5$.
2. Evaluate the function at the midpoint: Calculate $f(-1.5)$.
3. Choose the new interval:
   • If $f(-1.5)$ has the opposite sign of $f(-2)$, then the root lies in the interval [-2, -1.5].
   • If $f(-1.5)$ has the opposite sign of $f(-1)$, then the root lies in the interval [-1.5, -1].
4. Repeat: Go back to step 1 with our new, smaller interval.

Let's do the first iteration:

• Midpoint: -1.5
• $f(-1.5) = 4(-1.5)^3 + 2(-1.5)^2 - 6(-1.5) + 2 = -13.5 + 4.5 + 9 + 2 = 2$

Since $f(-1.5)$ is positive, and $f(-2)$ is negative, the root lies in the interval [-2, -1.5]. We've narrowed down our search!

Let's keep going. This is where a calculator or a computer really helps. We repeat the process until our interval is small enough that we're confident about the first two decimal places.

Iteration 2:

• Interval: [-2, -1.5]
• Midpoint: -1.75
• $f(-1.75) = 4(-1.75)^3 + 2(-1.75)^2 - 6(-1.75) + 2 hickapprox -3.8125$
• New Interval: [-1.75, -1.5] (because $f(-1.75)$ is negative and $f(-1.5)$ is positive)

Iteration 3:

• Interval: [-1.75, -1.5]
• Midpoint: -1.625
• $f(-1.625) = 4(-1.625)^3 + 2(-1.625)^2 - 6(-1.625) + 2 hickapprox -0.6621$
• New Interval: [-1.625, -1.5] (because $f(-1.625)$ is negative and $f(-1.5)$ is positive)

Iteration 4:

• Interval: [-1.625, -1.5]
• Midpoint: -1.5625
• $f(-1.5625) = 4(-1.5625)^3 + 2(-1.5625)^2 - 6(-1.5625) + 2 hickapprox 0.5994$
• New Interval: [-1.625, -1.5625] (because $f(-1.625)$ is negative and $f(-1.5625)$ is positive)

Iteration 5:

• Interval: [-1.625, -1.5625]
• Midpoint: -1.59375
• $f(-1.59375) = 4(-1.59375)^3 + 2(-1.59375)^2 - 6(-1.59375) + 2 hickapprox -0.0225$
• New Interval: [-1.59375, -1.5625] (because $f(-1.59375)$ is negative and $f(-1.5625)$ is positive)

Iteration 6:

• Interval: [-1.59375, -1.5625]
• Midpoint: -1.578125
• $f(-1.578125) = 4(-1.578125)^3 + 2(-1.578125)^2 - 6(-1.578125) + 2 hickapprox 0.2872$
• New Interval: [-1.59375, -1.578125] (because $f(-1.59375)$ is negative and $f(-1.578125)$ is positive)

Iteration 7:

• Interval: [-1.59375, -1.578125]
• Midpoint: -1.5859375
• $f(-1.5859375) = 4(-1.5859375)^3 + 2(-1.5859375)^2 - 6(-1.5859375) + 2 hickapprox 0.1322$
• New Interval: [-1.59375, -1.5859375] (because $f(-1.59375)$ is negative and $f(-1.5859375)$ is positive)

Iteration 8:

• Interval: [-1.59375, -1.5859375]
• Midpoint: -1.58984375
• $f(-1.58984375) = 4(-1.58984375)^3 + 2(-1.58984375)^2 - 6(-1.58984375) + 2 hickapprox 0.0547$
• New Interval: [-1.59375, -1.58984375] (because $f(-1.59375)$ is negative and $f(-1.58984375)$ is positive)

Iteration 9:

• Interval: [-1.59375, -1.58984375]
• Midpoint: -1.591796875
• $f(-1.591796875) = 4(-1.591796875)^3 + 2(-1.591796875)^2 - 6(-1.591796875) + 2 hickapprox 0.0161$
• New Interval: [-1.59375, -1.591796875] (because $f(-1.59375)$ is negative and $f(-1.591796875)$ is positive)

Iteration 10:

• Interval: [-1.59375, -1.591796875]
• Midpoint: -1.5927734375
• $f(-1.5927734375) = 4(-1.5927734375)^3 + 2(-1.5927734375)^2 - 6(-1.5927734375) + 2 hickapprox -0.0032$
• New Interval: [-1.5927734375, -1.591796875] (because $f(-1.5927734375)$ is negative and $f(-1.591796875)$ is positive)

After these iterations, our interval has shrunk considerably. We're trapped between -1.5927734375 and -1.591796875. To two decimal places, both of these numbers round to -1.59. So, we can confidently say that our solution, 'r', is approximately -1.59.

The Verdict: Our Approximate Solution

After all that work, we've arrived at our answer! The approximate solution to the equation $4x^3 + 2x^2 - 6x + 2 = 0$ in the interval [-2, -1], correct to two decimal places, is:

$$r hickapprox -1.59$$

Key Takeaways:

• Polynomial equations can have solutions that aren't whole numbers.
• The Intermediate Value Theorem helps us find intervals where roots exist.
• The bisection method is a reliable way to approximate roots by repeatedly narrowing down the interval.
• Approximations are a powerful tool in mathematics and are widely used in various applications.

So, there you have it! We've successfully navigated the world of root-finding and approximated a solution to a cubic equation. It might seem like a lot of steps, but each step is fairly simple, and the method guarantees that we'll get closer and closer to the true solution. Keep practicing, and you'll become a root-finding pro in no time!