Problem 57 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate. $$x^{2}+12 x=4$$

Step-by-Step Solution

Answer

Solutions: \( x = -6 + 2\sqrt{10} \) and \( x = -6 - 2\sqrt{10} \).

Step 1: Move All Terms to One Side

Start by moving all terms to one side of the equation to set it to zero. This gives us the standard form of a quadratic equation. Subtract 4 from both sides:\[ x^2 + 12x - 4 = 0 \]

Step 2: Use the Quadratic Formula

The quadratic formula is used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our equation, \( a = 1 \), \( b = 12 \), and \( c = -4 \).

Step 3: Calculate the Discriminant

Compute the discriminant \( b^2 - 4ac \):\[ b^2 - 4ac = 12^2 - 4 \cdot 1 \cdot (-4) = 144 + 16 = 160 \]

Step 4: Substitute into the Quadratic Formula

Plug the values into the quadratic formula:\[ x = \frac{-12 \pm \sqrt{160}}{2 \times 1} \]

Step 5: Simplify the Expression

Simplify the square root and the fraction:First, simplify \( \sqrt{160} \) to \( 4\sqrt{10} \):\[ \sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10} \]Then, simplify the expression:\[ x = \frac{-12 \pm 4\sqrt{10}}{2} \]Divide each term by 2:\[ x = -6 \pm 2\sqrt{10} \]

Step 6: State the Solutions

The solutions to the quadratic equation are:\[ x = -6 + 2\sqrt{10} \quad \text{and} \quad x = -6 - 2\sqrt{10} \]

Key Concepts

quadratic formula

The quadratic formula is a powerful tool used to find solutions to quadratic equations of the form \( ax^2 + bx + c = 0 \). It's especially handy when equations are not easily factorable. The quadratic formula is given by:

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

By using this formula, you can directly plug in the coefficients \( a \), \( b \), and \( c \) to find the value(s) of \( x \) that satisfy the equation. In our exercise, where we simplified the equation to \( x^2 + 12x - 4 = 0 \), we identified these coefficients as \( a = 1 \), \( b = 12 \), and \( c = -4 \).
Plugging these into the formula allows us to compute the potential solutions, making this method a comprehensive approach to solving almost any quadratic equation.

discriminant

The discriminant is part of the quadratic formula, represented by \( b^2 - 4ac \), and it gives us valuable information about the nature of the solutions of the quadratic equation. Depending on the value of the discriminant:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution, a repeated root.
• If \( b^2 - 4ac < 0 \), there are no real solutions, only complex ones.

In our example, we calculated the discriminant for the equation \( x^2 + 12x - 4 = 0 \) as \( 12^2 - 4 \times 1 \times (-4) = 160 \). Since 160 is greater than zero, this tells us we have two distinct real number solutions for \( x \). This critical aspect helps predict the kind of solutions to expect and guides us in our calculations.

standard form of quadratic equation

Quadratic equations must first be in their standard form to use the quadratic formula effectively. The standard form is expressed as \( ax^2 + bx + c = 0 \). This is essential because the formula relies on identifying the coefficients \( a \), \( b \), and \( c \).
In our exercise, the original equation \( x^2 + 12x = 4 \) needed to be rearranged to fit this form. We did this by subtracting 4 from each side, resulting in the equation \( x^2 + 12x - 4 = 0 \).
This is the standardized preparation for solving by various methods, particularly the quadratic formula. Ensuring the equation is set to zero allows us to systematically apply the mathematical tools available for finding solutions.

simplifying radical expressions

Simplifying radical expressions is a crucial step in obtaining the quadratic equation solutions in their simplest form. Often, quadratic formulas result in a square root that needs simplifying to make the solution easier to interpret and communicate.
For example, in the equation \( x^2 + 12x - 4 = 0 \), the square root derived as part of the quadratic formula was initially \( \sqrt{160} \).
This could be simplified by recognizing \( 160 = 16 \times 10 \), allowing us to break it down as \( \sqrt{16 \times 10} = 4\sqrt{10} \).
Simplifying radicals is about breaking down the number inside the square root into its prime factors, making it a neat calculation without losing accuracy in results. This not only makes calculations smoother but also helps in understanding the magnitude and nature of the solution terms more clearly.