Problem 84 - Elementary Algebra

Question

Solve using the quadratic formula. $$ -2(y-4)(y+1)=3 y+10 $$

Step-by-Step Solution

Answer

No real solutions exist as the discriminant is negative.

Step 1: Expand the Equation

First, expand the left side of the equation: \(-2(y-4)(y+1)\). Use the distributive property (FOIL method) to get:\[-2(y^2 + y - 4y - 4) = -2(y^2 - 3y - 4)\].Simplify it to get:\[-2y^2 + 6y + 8\].So the equation becomes:\(-2y^2 + 6y + 8 = 3y + 10\).

Step 2: Rearrange the Equation

Rearrange the equation to standard quadratic form (\(ax^2 + bx + c = 0\)):\(-2y^2 + 6y + 8 - 3y - 10 = 0\),which simplifies to:\(-2y^2 + 3y - 2 = 0\).

Step 3: Identify Coefficients

Identify the coefficients from the equation \(-2y^2 + 3y - 2 = 0\):- \(a = -2\)- \(b = 3\)- \(c = -2\).

Step 4: Apply the Quadratic Formula

Use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).Substitute the values:\[y = \frac{-3 \pm \sqrt{3^2 - 4(-2)(-2)}}{2(-2)}\].Calculate the discriminant:\[b^2 - 4ac = 3^2 - 4 \cdot (-2) \cdot (-2) = 9 - 16 = -7\].Since the discriminant is negative, this indicates that there are no real solutions.

Key Concepts

Distributive Property

The distributive property is a key tool in algebra when dealing with expressions inside parentheses. It allows you to multiply a term by all elements inside the parenthesis. This property is often expressed as \( a(b + c) = ab + ac \). In our original problem, you observe this property in action when expanding \(-2(y-4)(y+1)\). We use the distributive property—a process often referred to as the FOIL method when applied to binomials—to expand this expression fully. The steps include:

• Multiplying the first terms: \( y \times y = y^2 \).
• Multiplying the outer terms: \( y \times 1 = y \).
• Multiplying the inner terms: \(-4 \times y = -4y \).
• Multiplying the last terms: \(-4 \times 1 = -4 \).

Adding these together, you get \( y^2 - 3y - 4 \). Multiplying through by -2, we simplify the equation to \(-2y^2 + 6y + 8 \). Understanding this principle helps in reorganizing and simplifying polynomial expressions.

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. In our example, we rearranged the given equation to match this standard format. The given equation \(-2y^2 + 6y + 8 = 3y + 10 \) is first simplified and rearranged to \(-2y^2 + 3y - 2 = 0 \). The quadratic equation mainly involves:

• Identifying the coefficients: the \( a \) term represents the coefficient of \( y^2 \), \( b \) is the coefficient of \( y \), and \( c \) is the constant term.
• Locating these coefficients is critical for solving the equation using the quadratic formula or other methods.

Quadratic equations can yield up to two solutions, real or complex, depending on the discriminant's value, which we'll discuss next.

Discriminant

The discriminant helps determine the nature of the solutions in a quadratic equation, calculated by the expression \( b^2 - 4ac \). It reveals crucial information about the roots without requiring you to find them explicitly. Here’s how the discriminant is used:

• If \( b^2 - 4ac > 0 \), the equation has two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is exactly one real root, also known as a repeated root.
• If \( b^2 - 4ac < 0 \), there are no real roots, but two complex roots.

In the solution provided, the discriminant was calculated as \(-7\) (using \( 3^2 - 4(-2)(-2) \)), indicating there are no real solutions for the equation since \(-7\) is less than zero. Instead, there are complex solutions if you continue solving for \( y \). Understanding the discriminant helps predict the type of solutions a quadratic equation may have.