Problem 39 - Elementary Algebra

Question

Choose the appropriate method to solve the following. $$ x(x-6)=-29 $$

Step-by-Step Solution

Answer

The equation has solutions: \( x = 3 \pm 2\sqrt{5}i \).

Step 1: Expand the Quadratic Equation

To get the equation in standard quadratic form, expand the left-hand side of the equation:\[ x(x - 6) = x^2 - 6x \] This gives us the equation: \[ x^2 - 6x = -29 \]

Step 2: Move All Terms to One Side

Add 29 to both sides of the equation to get the quadratic in standard form:\[ x^2 - 6x + 29 = 0 \]

Step 3: Use the Quadratic Formula

The quadratic equation is in the form \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = -6 \), and \( c = 29 \). Use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute \( a \), \( b \), and \( c \) into the formula:\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times 29}}{2 \times 1} \]

Step 4: Calculate the Discriminant

Calculate the value of the discriminant \( b^2 - 4ac \):\[ (-6)^2 - 4 \times 1 \times 29 = 36 - 116 = -80 \]Since the discriminant is negative, the solutions are complex numbers.

Step 5: Compute the Solutions

Using the quadratic formula and the discriminant:\[ x = \frac{6 \pm \sqrt{-80}}{2} \]Simplify the square root of the negative number:\[ \sqrt{-80} = \sqrt{80}i = 4\sqrt{5}i \]Thus, the solutions are:\[ x = \frac{6 \pm 4\sqrt{5}i}{2} \]Further simplify:\[ x = 3 \pm 2\sqrt{5}i \]

Key Concepts

Quadratic Formula

The quadratic formula is an invaluable tool for finding the roots of any quadratic equation. A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The quadratic formula is given as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides the values of \( x \), which are the solutions to the equation. It requires calculating the discriminant, which is the expression under the square root: \( b^2 - 4ac \).
This discriminant helps determine the nature of the roots:

• If it is positive, the equation has two distinct real roots.
• If it is zero, the equation has one real root, repeated.
• If it is negative, the equation results in complex roots.

In the example, since the discriminant \( -80 \) is negative, the quadratic formula leads to complex number solutions.

Complex Numbers

Complex numbers extend the concept of regular numbers you are familiar with by introducing the imaginary unit \( i \), where \( i^2 = -1 \).

• A complex number is typically written as \( a + bi \), where \( a \) is the real part, and \( bi \) is the imaginary part.

These numbers are crucial when the solutions of quadratic equations involve negative discriminants. Solving an equation with a negative discriminant results in imaginary numbers in the form of \( bi \).
For example, the equation in the problem leads to a square root of \(-80\). Converting \( \sqrt{-80} \) using the imaginary unit gives \( \sqrt{80}i = 4\sqrt{5}i \).
Thus, showing the solutions are neither purely real nor imaginary but a blend of both real and imaginary parts, giving us complex solutions.

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is a key starting point for solving such equations. It is expressed in the format \( ax^2 + bx + c = 0 \). To solve a quadratic equation using the quadratic formula, the equation must first be in this standard form.
In the given exercise, the initial equation was \( x(x-6) = -29 \). This equation was expanded and reorganized to reach the standard form: \( x^2 - 6x + 29 = 0 \).
Breaking down the process:

• First, expand any products into terms: \( x(x - 6) \) expands to \( x^2 - 6x \).
• Then, ensure all terms are on one side, transforming any constants through addition or subtraction.
• This lets us easily identify suitable values of \( a \), \( b \), and \( c \) for solving with the quadratic formula.

Understanding this step is crucial, as correctly transforming your equation ensures effective use of advanced solutions like the quadratic formula.