Problem 52 - Intermediate Algebra

Question

Solve each quadratic equation using the method that seems most appropriate. $$ t(t-26)=-160 $$

Step-by-Step Solution

Answer

The solutions are \( t = 10 \) and \( t = 16 \).

Step 1: Expand the Given Equation

The given equation is \( t(t-26) = -160 \). Start by expanding it to form a standard quadratic equation: \( t^2 - 26t = -160 \).

Step 2: Bring All Terms to One Side

Rearrange the equation so that all terms are on one side and the equation is set to zero: \( t^2 - 26t + 160 = 0 \). This is our standard quadratic equation of the form \( ax^2 + bx + c = 0 \).

Step 3: Factor the Quadratic Equation

Next, factor the quadratic equation. We need to find two numbers that multiply to 160 and add to -26. These numbers are -10 and -16. Thus, we can factor the equation as \( (t - 10)(t - 16) = 0 \).

Step 4: Solve for t

Set each factor equal to zero to solve for \( t \). From \( t - 10 = 0 \), we get \( t = 10 \). From \( t - 16 = 0 \), we get \( t = 16 \). Thus, the solution is \( t = 10 \) or \( t = 16 \).

Key Concepts

Factoring Quadratics

Factoring is a fundamental technique for solving quadratic equations. A quadratic equation is in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In factoring, we express the quadratic equation as a product of two binomials. This simplifies the equation and allows us to solve for the variable easily.

• Firstly, identify two numbers that multiply to the constant term (\( c \)), and add up to the middle coefficient (\( b \)).
• Once the numbers are found, rewrite the quadratic equation in a factored form, such as \((x - m)(x - n) = 0 \), where \( m \) and \( n \) are the numbers found.

For example, consider the equation \( t^2 - 26t + 160 = 0 \). We look for two numbers that multiply to 160 and sum to -26. The numbers -10 and -16 fit perfectly. Thus, the factored form is \((t - 10)(t - 16) = 0 \).
Factoring is quick and effective when the equation is easily decomposable, making it a preferred method for simple quadratics.

Solving Quadratic Equations

Solving quadratic equations involves finding the value(s) of the variable that satisfy the equation. After factoring a quadratic equation, the next logical step is to solve it.

• Once the quadratic is factored, set each binomial equal to zero: \((x - m) = 0 \) and \((x - n) = 0 \).
• This gives us the solutions \( x = m \) and \( x = n \) respectively.

In our example, from the factored form \((t - 10)(t - 16) = 0 \), we set each factor equal to zero: \( t - 10 = 0 \) and \( t - 16 = 0 \). This yields the solutions \( t = 10 \) or \( t = 16 \).
This method provides clear and straightforward solutions, especially when the equation is factored successfully. Solving quadratic equations is foundational in algebra, as many real-world problems are modeled using quadratics.

Algebraic Methods

Algebraic methods encompass a variety of techniques for tackling quadratic equations. Besides factoring, these include using the quadratic formula and completing the square.

• Factoring: Simplifying the equation by expressing it as a product of two binomials.
• Quadratic Formula: Useful when factoring is complex or not possible, calculated as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Completing the Square: Involves rearranging terms to form a perfect square, thereby finding the solutions.

Each method has its own advantages depending on the equation's complexity and coefficients. Factoring is often the quickest if applicable, while the quadratic formula provides a surefire way to find solutions regardless of the equation's particularities.
For the equation \( t^2 - 26t + 160 = 0 \), factoring worked perfectly. However, if factoring were challenging, the quadratic formula could have easily provided solutions by substituting \( a = 1 \), \( b = -26 \), and \( c = 160 \) into the formula. This flexibility in methods provides a robust toolkit for solving any quadratic equation you may encounter.