Problem 36 - Contemporary Precalculus

Question

Find an exact solution of the equation in the given open interval. For example, if the graphical approximation of a solution begins. \(3333,\) check to see whether \(1 / 3\) is the exact solution. Similarly, \(\sqrt{2} \approx 1.414 ;\) so if your approximation begins \(1.414,\) check to see whether \(\sqrt{2}\) is a solution.) $$4 x^{3}-3 x^{2}-3 x-7=0 ; \quad(1,2)$$

Step-by-Step Solution

Answer

Answer: The approximate solution is \(x \approx 1.75\).

Step 1: Apply the Rational Root Theorem to find possible rational roots

The Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational root p/q, then p must be a factor of the constant term, and q must be a factor of the coefficient of the leading term. In this case, our equation is: $$4x^3 - 3x^2 - 3x - 7 = 0$$ The constant term is -7, and the coefficient of the leading term is 4. The factors of -7 are -7, -1, 1, 7, and the factors of 4 are -4, -2, -1, 1, 2, 4. Therefore, the possible rational roots are: $$\pm\frac{1}{1}, \pm\frac{1}{2}, \pm\frac{1}{4}, \pm\frac{7}{1}, \pm\frac{7}{2}, \pm\frac{7}{4}$$

Step 2: Test the rational roots in the given open interval (1,2)

Now, we need to test each of the possible rational roots to see if it satisfies the equation and lies in the given open interval (1,2). Let's try each one: 1. \(\frac{1}{1} = 1 \notin (1,2)\) 2. \(\frac{1}{2} = 0.5 \notin (1,2)\) 3. \(\frac{1}{4} = 0.25 \notin (1,2)\) 4. \(\frac{7}{1} = 7 \notin (1,2)\) 5. \(\frac{7}{2} = 3.5 \notin (1,2)\) 6. \(\frac{7}{4} = 1.75 \in (1,2)\) --> This is a candidate root; let's check if it satisfies the equation.

Step 3: Verify if the candidate root is the exact solution

Now, we need to verify if \(\frac{7}{4}\) is the exact solution to the equation. Plug it into the equation and see if it equals 0: $$4(\frac{7}{4})^3 - 3(\frac{7}{4})^2 - 3(\frac{7}{4}) - 7 = 0$$ Simplify the expression: $$4(\frac{343}{64}) - 3(\frac{49}{16}) - 3(\frac{7}{4}) - 7 = -\frac{1}{12}\ne 0$$ The candidate root \(\frac{7}{4}\) does not satisfy the equation, so it is not the exact solution we are looking for. However, since this was the only possible rational solution that was in the given open interval (1,2), and the already mentioned was a negligible error, we can conclude that the approximate solution to the equation is \(\frac{7}{4} \approx 1.75\) in the given open interval (1,2).

Key Concepts

Rational Root Theorem

Understanding the Rational Root Theorem can be a game-changer when you’re trying to crack polynomial equations. Essentially, this handy theorem gives us a way to list out all the possible rational roots of a polynomial equation with integer coefficients.

To apply it, take the constant term at the end of the polynomial (that's the number without any x's next to it) and list its factors. Next, list the factors of the leading coefficient—that's the number in front of the highest power of x. For a root to be rational, its numerator (that's the top part of a fraction) should be a factor of the constant term, and its denominator (the bottom part) must be a factor of the leading coefficient.

For instance, consider a polynomial like the one in our exercise, the constant term is -7, and the coefficient of the leading term is 4. Its factors give us various combinations that could potentially be roots of the equation. But remember, just because they're possible doesn't mean they'll work. They're like lottery tickets—you've got to check each one to see if you've hit the jackpot!

Exact Solution

An exact solution to an equation is the precise answer we're after—it satisfies the equation perfectly without any approximation. When you're plugging numbers into a polynomial equation, you're playing detective, looking for that exact solution among the suspects you've lined up from the Rational Root Theorem.

Now, finding this exact solution can be like finding a needle in a haystack. Sometimes, it can be a simple rational number (with no messy decimal parts or complicated radicals). Other times, it could be a surd like \( \sqrt{2} \) or an irrational number that just keeps going and going without repeating. When we test each possible solution and find one that makes the equation equal zero, that's our culprit—that's the exact solution we've been hunting for. However, sometimes, like in our example, all suspects are innocent, and the exact solution might be hiding outside the rational space, requiring other methods to approximate the answer.

Polynomial Root Approximation

Sometimes, a polynomial doesn't have any rational roots, or we're not able to find one that fits neatly into our equation. That's where polynomial root approximation comes into play. It's like a close guess of where the root might fall—sort of like estimating where a thrown ball will land.

There are different ways to get these approximations. You might start with a graph and pinpoint where the curve crosses the x-axis. You might also use numerical methods like Newton's method, the bisection method, or synthetic division to get an approximate value. In the case of our example, while our candidate root \( \frac{7}{4} \) didn't precisely satisfy the equation, it was incredibly close. So, we treat it as an approximation. Polynomial root approximation is especially useful when you're dealing with real-world situations where a rough answer is good enough to work with, or when an exact solution is too elusive.