Problem 37 - Algebra A Combined Function

Question

Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. $$ \frac{1}{2} y^{2}=y-\frac{1}{2} $$

Step-by-Step Solution

Answer

The solution is \( y = 1 \), a repeated real root.

Step 1: Rewrite the equation in standard form

The given equation is \( \frac{1}{2} y^{2} = y - \frac{1}{2} \). Before applying the quadratic formula, it is important to rearrange it to the standard form \( ay^{2} + by + c = 0 \). Multiply every term by 2 to get \( y^{2} = 2y - 1 \). Move all terms to one side to get the equation \( y^{2} - 2y + 1 = 0 \).

Step 2: Identify coefficients a, b, and c

For the equation \( y^{2} - 2y + 1 = 0 \), identify coefficients: \( a = 1 \), \( b = -2 \), and \( c = 1 \).

Step 3: Apply the quadratic formula

The quadratic formula is \( y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = -2 \), and \( c = 1 \) into the formula.

Step 4: Calculate the discriminant

First, calculate the discriminant \( b^{2} - 4ac \). Here \( b^2 = (-2)^2 = 4 \) and \( 4ac = 4(1)(1) = 4 \). So, \( b^{2} - 4ac = 4 - 4 = 0 \).

Step 5: Solve for y using the quadratic formula

Since the discriminant is 0, this means there is exactly one real solution (a repeated root). Substitute the discriminant and other values into the quadratic formula: \( y = \frac{-(-2) \pm \sqrt{0}}{2 \times 1} = \frac{2}{2} = 1 \). So, the solution is \( y = 1 \).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra and can be expressed in the standard form: \[ ax^2 + bx + c = 0 \]. Here, \( a \), \( b \), and \( c \) represent constants, where \( a eq 0 \). The variable \( x \) is what we're solving for, known as the unknown or variable. Every quadratic equation can graphically be represented as a parabola, a U-shaped curve, on a coordinate plane.

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

Quadratic equations are notably solved via various methods such as factoring, completing the square, or using the quadratic formula. Using the formula is especially handy when the equation does not easily factor or when you want to find exact solutions.

Discriminant

The discriminant is a key component in the quadratic formula, denoted as \( b^2 - 4ac \). It provides crucial information about the nature and number of the solutions of a quadratic equation. By examining the discriminant:

• If \( b^2 - 4ac > 0 \), the equation has two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution, often referred to as a repeated or double root.
• If \( b^2 - 4ac < 0 \), the solutions are non-real (complex) and occur as complex conjugates.

For the equation provided in the example, the discriminant is 0, indicating that there is a single repeated real solution. Calculating the discriminant is an efficient way to gauge the solvability of a quadratic equation before fully solving it.

Real Solutions

Real solutions of quadratic equations are those that can be plotted on the real number line. When solving a quadratic equation using the quadratic formula, real solutions occur when the discriminant \( b^2 - 4ac \) is zero or positive. In our example, the discriminant was found to be 0, which implies a real solution. The math behind it shows that the quadratic equation reduces to having a single solution, what we call a repeated root. This means that the point on the parabola just touches the x-axis at this one value.For the example equation, the real solution is \( y = 1 \), and this shows that the parabola touches the x-axis at this single point, emphasizing the concept of repeated roots.

Complex Solutions

Complex solutions arise when the discriminant \( b^2 - 4ac \) of a quadratic equation is negative. In such cases, the solutions include imaginary numbers, indicated because the square root of a negative number involves the imaginary unit \( i \). Mathematically, when you solve such equations, the solutions take the form:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( \sqrt{b^2 - 4ac} \) becomes an imaginary number because of the negative discriminant.These solutions, often referred to as complex conjugates, come in pairs like \( a + bi \) and \( a - bi \). While they don't correspond to any point on the real number line, they provide essential information about the properties of a quadratic graph. For instance, they can denote how a parabola behaves in relation to the x-axis, such as not intersecting it at any point. However, in our given example, complex solutions don't occur as the discriminant was zero, resulting in real solutions.