Problem 15 - Algebra for College Students

Question

Use the quadratic formula to solve each of the quadratic equations. Check your solutions by using the sum and product relationships. $$a^{2}-8 a=4$$

Step-by-Step Solution

Answer

The solutions are \(a = 4 \pm 2\sqrt{5}\).

Step 1: Write the Equation in Standard Form

First, we need to rewrite the given equation in the standard quadratic form, which is \(ax^2 + bx + c = 0\). The given equation is \(a^2 - 8a = 4\). To convert it, subtract 4 from both sides to get \(a^2 - 8a - 4 = 0\).

Step 2: Identify Coefficients

In the quadratic equation \(a^2 - 8a - 4 = 0\), identify the coefficients needed for the quadratic formula. Here, \(a = 1\), \(b = -8\), and \(c = -4\).

Step 3: Use the Quadratic Formula

The quadratic formula is \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting the coefficients from our equation, we have \(a = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-4)}}{2(1)}\).

Step 4: Simplify the Expression

Calculate the parts of the formula: \(-(-8) = 8\), and \((-8)^2 = 64\). Then, calculate the discriminant: \(64 + 16 = 80\). So the equation becomes \(a = \frac{8 \pm \sqrt{80}}{2}\).

Step 5: Simplify the Solution

Since \(\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}\), substitute back to get \(a = \frac{8 \pm 4\sqrt{5}}{2}\). Thus, \(a_1 = 4 + 2\sqrt{5}\) and \(a_2 = 4 - 2\sqrt{5}\).

Step 6: Check Solutions with Sum and Product

The sum of the solutions is \(a_1 + a_2 = (4 + 2\sqrt{5}) + (4 - 2\sqrt{5}) = 8\), which equals \(-b/a\). The product of the solutions is \(a_1 \, a_2 = (4 + 2\sqrt{5})(4 - 2\sqrt{5}) = 16 - 20 = -4\), which equals \(c/a\). The checks confirm the solutions are correct.

Key Concepts

Quadratic Equation

A quadratic equation is an equation of the second degree, which means it includes at least one term that is squared. The general structure of a quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) represents an unknown variable. This type of equation is fundamental in algebra and appears in various real-life contexts, such as physics, engineering, and finance. Quadratic equations can have 0, 1, or 2 real solutions, and finding these solutions is crucial for drawing graphs or solving optimization problems.

Standard Form

The standard form of a quadratic equation is a way of writing the equation as \(ax^2 + bx + c = 0\). This format is beneficial because it sets the stage for using powerful techniques, like factoring, completing the square, or using the quadratic formula to solve the equation.
In the given exercise, the equation \(a^2 - 8a = 4\) is not initially in standard form. To convert it, you must rearrange it to \(a^2 - 8a - 4 = 0\). This involves moving all terms to one side of the equation to comply with the standard format. Once this arrangement is complete, it becomes straightforward to identify the coefficients \(a = 1\), \(b = -8\), and \(c = -4\), which are crucial for solving the quadratic equation using the quadratic formula.

Discriminant

The discriminant in a quadratic equation is a component of the quadratic formula that lies under the square root: \(b^2 - 4ac\). It provides essential information about the roots of the equation without solving it completely. The value of the discriminant determines the nature of the roots:

• If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
• If \(b^2 - 4ac = 0\), the equation has one real root (or a repeated root).
• If \(b^2 - 4ac < 0\), the equation has no real roots (instead, it has two complex roots).

In the current problem, we find the discriminant by calculating \((-8)^2 - 4(1)(-4)\), which results in \(80\). Because \(80 > 0\), this tells us that our equation indeed has two distinct real roots, corresponding to the solutions obtained from the quadratic formula.

Sum and Product of Roots

The sum and product of roots of a quadratic equation are useful properties that provide a way to verify the correctness of your solutions without extensive computation. These are derived from Vieta's formulas:

• The sum of the roots \(x_1 + x_2\) is given by \(-\frac{b}{a}\).
• The product of the roots \(x_1 \cdot x_2\) is given by \(\frac{c}{a}\).

In the exercise, the solutions found were \(a_1 = 4 + 2\sqrt{5}\) and \(a_2 = 4 - 2\sqrt{5}\). Their sum is \(8\), which matches with \(-\frac{-8}{1} = 8\). Similarly, their product is \(-4\), confirming \(\frac{-4}{1} = -4\). Such checks provide an excellent way to ensure that your solution is correct and robust against small calculation errors.