Problem 20 - Algebra A Combined Function

Question

Use the quadratic formula to solve each equation. These equations have real number solutions only. $$ y^{2}-8=4 y $$

Step-by-Step Solution

Answer

The solutions are \( y = 2 + 2\sqrt{3} \) and \( y = 2 - 2\sqrt{3} \).

Step 1: Rewrite the Equation

The given equation is \( y^2 - 8 = 4y \). Begin by moving all terms to one side of the equation to set it to zero. This will yield \( y^2 - 4y - 8 = 0 \).

Step 2: Identify Coefficients

The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = -4 \), and \( c = -8 \).

Step 3: Apply the Quadratic Formula

The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the identified coefficients: \( y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \).

Step 4: Calculate the Discriminant

The discriminant \( b^2 - 4ac \) is \( (-4)^2 - 4 \times 1 \times (-8) = 16 + 32 = 48 \). Since the discriminant is positive, there are two real solutions.

Step 5: Solve for y

Substitute the discriminant back into the quadratic formula: \( y = \frac{4 \pm \sqrt{48}}{2} \). Simplify \( \sqrt{48} \) to \( 4\sqrt{3} \), giving \( y = \frac{4 \pm 4\sqrt{3}}{2} \).

Step 6: Simplify the Solutions

Divide both terms in the numerator by 2: \( y = 2 \pm 2\sqrt{3} \). Thus, the solutions are \( y = 2 + 2\sqrt{3} \) and \( y = 2 - 2\sqrt{3} \).

Key Concepts

Quadratic Equation

A quadratic equation is any equation that can be rearranged in standard form as \( ax^2 + bx + c = 0 \). This type of equation represents a parabola in the coordinate system and involves variable terms where the highest exponent is two. The letters \(a\), \(b\), and \(c\) are coefficients, where

• \(a\) is the coefficient of \(x^2\), and it must not be zero (otherwise, it is not quadratic).
• \(b\) is the coefficient of \(x\).
• \(c\) is the constant term.

Understanding how to approach these equations involves recognizing their structure and manipulating them to one standard form. In the case given, the equation \( y^2 - 8 = 4y \) was rewritten as \( y^2 - 4y - 8 = 0 \). Recognizing that \(a = 1\), \(b = -4\), and \(c = -8\) allows us to solve it using designated methods like factoring, completing the square, or, as used here, the quadratic formula.

Discriminant

The discriminant is a component of the quadratic formula that helps to determine the nature and number of the roots of a quadratic equation. It is found within the formula as \( b^2 - 4ac \). This value can tell you if the solutions to a quadratic equation are:

• Two distinct real numbers if the discriminant is positive.
• One unique real number if the discriminant is zero.
• Two complex numbers if the discriminant is negative.

In the provided scenario, the discriminant is \( 48 \), which is determined in Step 4 of the solution. Calculating \( (-4)^2 - 4 \times 1 \times (-8) = 16 + 32 = 48 \), you can see it is positive. Therefore, we know this quadratic equation has two distinct real solutions, and this guides how we proceed with solving for \( y \), ensuring we calculate both possible outcomes.

Real Solutions

The solutions to a quadratic equation, known as roots, can either be real or complex, and are found using methods like the quadratic formula. In our specific example, after identifying the discriminant as positive, we can confirm two real solutions exist. These are computed using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Here, substituting the known values provides two calculations due to the "\( \pm \)":

• \( y = \frac{4 + 4\sqrt{3}}{2} \)
• \( y = \frac{4 - 4\sqrt{3}}{2} \)

These simplify to \( y = 2 + 2\sqrt{3} \) and \( y = 2 - 2\sqrt{3} \), respectively. The roots \( y = 2 + 2\sqrt{3} \) and \( y = 2 - 2\sqrt{3} \) are the points where the parabola intersects the vertical line in the coordinate system, offering solutions in real numbers and making this clear through their simplicity.