Problem 10 - Basic Technical Mathematics with Calculus

Question

Solve the given equations without using a calculator. $$4 x^{3}-16 x^{2}+21 x-9=0$$

Step-by-Step Solution

Answer

The solutions are x = 1 and x = \(\frac{3}{2}\).

Step 1: Factor the Equation or Use the Rational Root Theorem

The Rational Root Theorem suggests that possible rational roots are factors of the constant term, -9, divided by factors of the leading coefficient, 4. Possible rational roots are ±1, ±3, ±9, ±1/2, ±3/2, ±9/2, ±1/4, ±3/4, ±9/4.

Step 2: Test Possible Roots

Test these possible roots by substituting into the equation to see if they equal zero. Start with simple ones like x = 1, x = -1, x = 3, etc. Substitute x = 3:\[ 4(3)^3 - 16(3)^2 + 21(3) - 9 = 108 - 144 + 63 - 9 = 18 \]Since this does not equal zero, x = 3 is not a root. Test other values similarly.

Step 3: Identify a Root

Upon substituting x = 1 into the equation:\[ 4(1)^3 - 16(1)^2 + 21(1) - 9 = 4 - 16 + 21 - 9 = 0 \]Thus, x = 1 is a root.

Step 4: Factor Out the Root

Since x = 1 is a root, divide the polynomial by (x - 1) using synthetic division or polynomial division: Performing synthetic division with x = 1: 4 -16 21 -9 1 | 4 -12 9 0 -------------------- 4 -12 9 Result: 4x² - 12x + 9.

Step 5: Solve the Quadratic Equation

Now solve the quadratic equation \(4x^2 - 12x + 9 = 0\) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a = 4\), \(b = -12\), and \(c = 9\):\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(4)(9)}}{2(4)} = \frac{12 \pm \sqrt{144 - 144}}{8} = \frac{12 \pm 0}{8} = \frac{12}{8} = \frac{3}{2} \]Thus, the double root is x = \(\frac{3}{2}\).

Step 6: Identify All Solutions

The solutions to the equation are the root found in Step 3 and the roots of the quadratic from Step 5. These are x = 1 and x = \(\frac{3}{2}\).

Key Concepts

Polynomial Equations

Polynomial equations are mathematical statements that involve polynomials. A polynomial is an expression that can have constants, variables, and exponents. A typical polynomial looks like this:

• The exponents on the variables are whole numbers.
• There's no division by a variable.
• They usually have a finite number of terms.

For instance, the polynomial equation given in the exercise is \(4x^3 - 16x^2 + 21x - 9 = 0\). The goal is to find values of \(x\) that make the equation true, which is known as finding the roots of the equation.
These roots can be real or complex numbers. In our example, we focus on finding real roots, which are the solutions you would typically expect to see as intersections with the x-axis when the polynomial is graphed.

Rational Root Theorem

The Rational Root Theorem is a helpful tool in algebra for finding possible rational solutions to polynomial equations. It provides a list of potential rational roots based on the polynomial's coefficients.
This theorem states that any possible rational root, if it exists, is a factor of the constant term divided by a factor of the leading coefficient.

• In our polynomial \(4x^3 - 16x^2 + 21x - 9\), the constant term is \(-9\), and the leading coefficient is \(4\).
• Therefore, potential rational roots are factors of \(-9\) (like \(±1, ±3, ±9\)) divided by factors of \(4\) (like \(±1, ±2, ±4\)).

By testing these possibilities, we can identify zeroes of the polynomial without excessive calculation. This application helps reduce the workload before using other methods like synthetic division.

Synthetic Division

Synthetic division is a shortcut method to divide polynomials. It's simpler than long division when dealing with polynomials and is particularly useful for dividing by a linear factor like \(x - c\).
When you know a root like \(x = 1\) for a polynomial \(4x^3 - 16x^2 + 21x - 9\), you can use it to divide the polynomial by the root's linear factor, \(x - 1\).
After performing synthetic division, the polynomial reduces to \(4x^2 - 12x + 9\), which is easier to solve. The steps involve:

• Writing down the coefficients of the polynomial.
• Using the known root, perform synthetic division.
• Arrive at the reduced polynomial.

This makes further solutions, whether by factoring or using formulas, more manageable.

Quadratic Formula

The quadratic formula is a classic way to solve quadratic equations, which are of the form \(ax^2 + bx + c = 0\). It works for all quadratic equations, even when other factoring methods don't.
For the reduced equation \(4x^2 - 12x + 9 = 0\), we use this formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 4\), \(b = -12\), and \(c = 9\). Plugging in these values, we solve:

• The term \(b^2 - 4ac\) under the square root, known as the discriminant, is zero in this example.
• The result is a single solution \(x = \frac{3}{2}\), known as a repeated or double root since the discriminant is zero.

This formula is powerful and ensures you find all possible solutions to a quadratic, making it an essential tool in algebra.