Problem 49 - Algebra and Trigonometry with Analytic Geometry

Question

Find the solutions of the equation $$ 27 x^{3}=(x+5)^{3} $$

Step-by-Step Solution

Answer

The solutions are \( x = -1, 1.5769, -0.7692 \).

Step 1: Set Equation to Standard Form

Start by expanding the expression on the right side of the equation. The equation is \( 27x^3 = (x+5)^3 \), which expands to \( 27x^3 = x^3 + 3\cdot x^2 \cdot 5 + 3 \cdot x \cdot 5^2 + 5^3 \). After simplifying, you get \( 27x^3 = x^3 + 15x^2 + 75x + 125 \).

Step 2: Rearrange the Equation

Next, set the equation to zero by moving all terms to one side: \( 27x^3 - (x^3 + 15x^2 + 75x + 125) = 0 \).

Step 3: Simplify the Equation

Combine like terms from the rearranged equation: \( 27x^3 - x^3 - 15x^2 - 75x - 125 = 0 \). This simplifies to \( 26x^3 - 15x^2 - 75x - 125 = 0 \).

Step 4: Factor the Equation

Notice that the equation can be factored by extracting common factors or by trial and error. In this case, use trial values of \( x \) to find a root.

Step 5: Find a Root Through Trial and Error

Try \( x = 1 \): substituting into the equation gives \( 26(1)^3 - 15(1)^2 - 75(1) - 125 = -189 \), which is not zero. Next, try \( x = -1 \): \( 26(-1)^3 - 15(-1)^2 - 75(-1) - 125 = 0 \). So, \( x = -1 \) is a root.

Step 6: Use Synthetic Division

Divide the polynomial \( 26x^3 - 15x^2 - 75x - 125 \) by \( x + 1 \) using synthetic division to find the quotient: \( 26x^2 - 41x - 125 \).

Step 7: Solve the Quadratic Equation

Solve \( 26x^2 - 41x - 125 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 26, b = -41, c = -125 \).

Step 8: Calculate the Roots

Calculate the discriminant: \( b^2 - 4ac = (-41)^2 - 4\cdot26\cdot(-125) = 1681 \). Compute the roots: \( x = \frac{41 \pm \sqrt{1681}}{52} \), which gives \( x = \frac{41 \pm 41}{52} \). The solutions are \( x = 1.5769 \) and \( x = -0.7692 \).

Step 9: Compile the Complete Solution Set

Including the real root found earlier, the solutions to the equation are \( x = -1, 1.5769, -0.7692 \).

Key Concepts

Cubic Equations

Cubic equations are a type of polynomial equation where the highest exponent of the variable, usually represented as \(x\), is 3. These equations typically have the form \(ax^3 + bx^2 + cx + d = 0\), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(a eq 0\). Solving cubic equations is more complex than solving linear or quadratic equations, primarily due to the additional degree.Cubic equations can have up to three real roots, one of which may be complex if the discriminant of the equation is negative. The process of solving these equations often involves reducing the equation gradually- first by isolating terms and then using methods such as factoring or synthetic division.For example, in our exercise, the equation \(27x^3 = (x+5)^3\) was expanded and rearranged into standard form, yielding \(26x^3 - 15x^2 - 75x - 125 = 0\). The goal was to manipulate the equation to find its real roots and then further solve any remaining factored or simplified parts to identify additional solutions.

Synthetic Division

Synthetic division is a simplified form of polynomial division, specifically used for dividing a polynomial by a linear factor of the form \(x - c\). It's an efficient way to divide without having to go through the long process of traditional polynomial division.When we apply synthetic division, we focus on the coefficients of the polynomial. For example, in our solution, after identifying \(x = -1\) as a root through trial and error, the polynomial \(26x^3 - 15x^2 - 75x - 125\) was divided by \(x + 1\). The coefficients \(26, -15, -75, -125\) were used, and the synthetic division process helped simplify the polynomial into \(26x^2 - 41x - 125\).Here's a simple outline of synthetic division:

• Write down the coefficients of the polynomial.
• Use the root value to perform the division across the coefficients.
• After you run through the coefficients, the result will give you the quotient polynomial, which is one degree lower than the original.

This process streamlines the task of finding other polynomial roots once one root is known.

Quadratic Formula

The quadratic formula provides a solution for finding the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). It is specifically useful when the quadratic equation cannot be easily factored.The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a\), \(b\), and \(c\) represent the coefficients of the quadratic equation.In the context of our exercise, once the cubic equation was reduced to a quadratic form, \(26x^2 - 41x - 125 = 0\), we applied the quadratic formula. The discriminant \(b^2 - 4ac\) was calculated to determine the nature of the roots:

• If the discriminant is positive, two distinct real solutions exist.
• If it is zero, only one real solution is present, indicating a repeated root.
• Negative discriminants imply complex roots.

By plugging into the formula, we obtained the solutions \(x = 1.5769\) and \(x = -0.7692\), completing the solution alongside the already discovered root \(x = -1\).