Problem 56 - Introductory Algebra for College Students

Question

Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$25 w^{2}=80 w-64$$

Step-by-Step Solution

Answer

The solutions to this quadratic equation are \(w = 4/5\) and \(w = 16/5\).

Step 1: Rewrite the equation in standard form

We start this problem by putting the quadratic equation in a standard form. This is done by moving everything to one side to equal to zero: \(25w^2 - 80w + 64 = 0\).

Step 2: Factor the quadratic equation

The quadratic equation is factorable if there exist two numbers that add to give -80 (coefficient of w) and multiply to give 64 (constant term). After some trial and error or by using quadratic formula, you can find these two numbers are -4 and -16. So, the factored form would be \((5w - 4)(5w - 16) = 0\).

Step 3: Solve for w

After factoring, solve the equation by setting each factor equal to 0 and solving for w. You'll get two possible solutions \(w1 = 4/5\) and \(w2 = 16/5\).

Step 4: Check the solutions

To verify the solutions, substitute each solution to the original equation and see if it makes the equation true. For \(w1 = 4/5\), the original equation becomes \(25 * (4/5)^2 = 80 * (4/5) - 64\), which simplifies to 16 = 16, a true statement. Therefore \(w1 = 4/5\) is a solution. Similarly, for \(w2 = 16/5\), the original equation becomes \(25 * (16/5)^2 = 80 * (16/5) - 64\), which simplifies to 256 = 256, another true statement. Therefore \(w2 = 16/5\) is also a solution.

Key Concepts

Factoring

Factoring is a useful method for solving quadratic equations. Imagine you need to break down an equation, like a tricky puzzle, into simpler pieces. This process involves expressing the quadratic in the form

• \(ax^2 + bx + c = 0\)
• after finding two binomials \((px + q)(rx + s)\) that multiply out to the original equation.

Start by identifying factors of the quadratic's leading coefficient and constant term. For example, with the equation \(25w^2 - 80w + 64 = 0\), look for numbers that add up to \(-80\) and multiply to \(64\). After some trials, you find \(-4\) and \(-16\), giving you the factored form \((5w - 4)(5w - 16) = 0\).
Now, solve for \(w\) by setting each factor equal to zero, resulting in solutions for \(w\).

Quadratic Formula

The quadratic formula is like a magic key that can always unlock solutions for any quadratic equation. It is given by:

• \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

For any quadratic in the form \(ax^2 + bx + c = 0\), just plug in the coefficients \(a\), \(b\), and \(c\), and the formula does the rest. This is especially helpful if factoring is difficult or impossible.
In the problem \(25w^2 - 80w + 64 = 0\), although we factored it, using the quadratic formula would equally give us the same solutions: \(w = 4/5\) and \(w = 16/5\). Substitute and simplify to find your solutions. It's always a reliable option!

X-Intercepts

Finding x-intercepts means discovering the values where a graph crosses the x-axis. For any quadratic equation \(ax^2 + bx + c\), the solutions (or roots) represent the x-intercepts.
Think of x-intercepts as spots where the height (y-value) is zero. This happens when you solve the equation \(ax^2 + bx + c = 0\). For the equation \(25w^2 - 80w + 64 = 0\), solving gives us \(w = 4/5\) and \(w = 16/5\).
On the graph, the curve will touch or cross the x-axis at these points. You can use graphing tools to visualize this, providing a clear picture of how everything connects. Remember, finding x-intercepts helps confirm your solutions and better understand the graph's behavior.
It's a handy way to check your work!