Problem 54 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ 2 x^{2}-5 x=-2 $$

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = \frac{1}{2}\).

Step 1: Move all terms to one side of the equation

To solve the quadratic equation \(2x^2 - 5x = -2\), start by moving all terms to one side of the equation to set it equal to zero. Add 2 to both sides to get: \[ 2x^2 - 5x + 2 = 0 \]

Step 2: Factor the quadratic equation

Next, we need to factor the quadratic equation \(2x^2 - 5x + 2\). We look for two numbers that multiply to \(2 \times 2 = 4\) and add to \(-5\). The numbers \(-1\) and \(-4\) meet these criteria. We can use these to rewrite the middle term: \[ 2x^2 - x - 4x + 2 = 0 \]

Step 3: Group terms and factor by grouping

Group the terms in pairs: \[(2x^2 - x) - (4x - 2) = 0\] Factor out the greatest common factor in each group: \[x(2x - 1) - 2(2x - 1) = 0\] Notice that \(2x - 1\) is a common factor.

Step 4: Factor out the common binomial factor

Now, factor out the common factor \(2x - 1\) from the equation: \[(x - 2)(2x - 1) = 0\]

Step 5: Solve each factor for x

To find the solutions, set each factor equal to zero and solve:1. \(x - 2 = 0\) Solve for \(x\): \(x = 2\)2. \(2x - 1 = 0\) Add 1 to both sides: \(2x = 1\) Divide both sides by 2: \(x = \frac{1}{2}\)

Step 6: State the final solution

The solutions to the equation \(2x^2 - 5x + 2 = 0\) are \(x = 2\) and \(x = \frac{1}{2}\).

Key Concepts

Factoring in Quadratic Equations

Factoring is a crucial skill for solving quadratic equations because it allows us to break down the equation into simpler, more manageable parts. Think of a quadratic equation as a mystery box containing two hidden numbers that multiply to give a constant number (the last term) and add to result in the linear coefficient (the middle term). By factoring, we're essentially unpacking these hidden factors.
Let's consider the equation: \[2x^2 - 5x + 2 = 0\] The task is to identify two numbers that multiply to the product of the leading coefficient (2) and the constant term (2) which is 4, but also add up to the linear coefficient, -5.

• We found -1 and -4 as our key players since they fit the criteria.
• We then split the middle term -5x into -x and -4x, forming smaller groups:

\[(2x^2 - x) - (4x - 2) = 0\]This process prepares each group for further factoring, crucially leading us to a factored form where solving becomes straightforward.

Understanding the Quadratic Formula

The quadratic formula is a universal tool in our toolkit—capable of solving any quadratic equation. Despite using factoring above, it's essential to know how versatile and reliable this method is when factoring becomes complex or impossible. The formula is expressed as follows:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• For any quadratic equation in the form of \(ax^2 + bx + c = 0\), the coefficients are plugged into the formula to find the roots.
• The term under the square root, \(b^2 - 4ac\), is called the discriminant, which gives insight into the nature of the roots.

For our specific equation, substituting in the values (with \(a = 2\), \(b = -5\), and \(c = 2\)), would provide the solutions directly. While this tool wasn't needed in this case due to successful factoring, it's reassuring to know there's a failsafe approach when other methods fall short.

Applying the Zero Product Property

The zero product property is a simple yet powerful idea: if two numbers multiply to zero, at least one of them must be zero. This principle is a cornerstone in solving equations once they have been factored.In our example, after factoring, we end up with:\[(x - 2)(2x - 1) = 0\]

• The zero product property tells us to set each factor individually equal to zero:
• \(x - 2 = 0\) leads to \(x = 2\)
• \(2x - 1 = 0\) simplifies to \(x = \frac{1}{2}\)

Each factor holds the potential to solve the entire equation. This is why factoring is so useful—it allows us to apply the zero product property, leading us to each potential solution quickly and accurately.