Problem 69 - A Graphical Approach to Precalculus with Limits

Question

Solve each quadratic equation by completing the square. $$2 x(2 x-5)=2$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{5 + \sqrt{33}}{4}\) and \(x = \frac{5 - \sqrt{33}}{4}\).

Step 1: Expand and Simplify the Quadratic Equation

Start by expanding and rearranging the given equation. The original equation is:\[2x(2x-5) = 2\]First, distribute 2x:\[4x^2 - 10x = 2\]Then, move 2 to the left side of the equation:\[4x^2 - 10x - 2 = 0\]

Step 2: Divide by the Leading Coefficient

To make completing the square easier, divide every term by the leading coefficient (4):\[x^2 - \frac{5}{2}x - \frac{1}{2} = 0\]

Step 3: Move the Constant Term to the Right Side

Rearrange the equation to move the constant term to the right side:\[x^2 - \frac{5}{2}x = \frac{1}{2}\]

Step 4: Complete the Square

To complete the square, add and subtract the square of half the coefficient of x. Half of \(-\frac{5}{2}\) is \(-\frac{5}{4}\), and its square is \(\left(-\frac{5}{4}\right)^2 = \frac{25}{16}\). Add \(\frac{25}{16}\) to both sides:\[x^2 - \frac{5}{2}x + \frac{25}{16} = \frac{1}{2} + \frac{25}{16}\]The left side becomes a perfect square:\[(x - \frac{5}{4})^2 = \frac{8}{16} + \frac{25}{16}\]

Step 5: Simplify the Right Side

Combine the fractions on the right side:\[(x - \frac{5}{4})^2 = \frac{33}{16}\]

Step 6: Solve for x by Taking the Square Root

Take the square root of both sides, considering both the positive and negative roots:\[x - \frac{5}{4} = \pm \frac{\sqrt{33}}{4}\]

Step 7: Solve for x

Add \(\frac{5}{4}\) to both sides to solve for x:\[x = \frac{5}{4} \pm \frac{\sqrt{33}}{4}\]

Step 8: Write the Final Solutions

The solutions for x are:\[x = \frac{5 + \sqrt{33}}{4}\] and \[x = \frac{5 - \sqrt{33}}{4}\]

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic equation into a form that is easier to solve. The goal is to reshape the quadratic equation in such a way that one side forms a complete square trinomial. This can be particularly useful because once in this form, it's simple to solve by taking the square root.

Here's how it works:

• Start with a quadratic equation in the form of \(ax^2 + bx + c = 0\).
• Move the constant term, \(c\), to the other side of the equation to allow a clear space to complete the square.
• If the coefficient \(a\) in front of \(x^2\) isn't 1, divide the entire equation by \(a\) to simplify the work.
• Now, focus on the \(x^2\) and \(x\) terms. Take half of the coefficient of the \(x\) term, square it, and add this square to both sides of the equation. This completes the square on one side.
• The equation is then of the form \((x + d)^2 = e\), where \(d\) is half the original \(b\) coefficient (from the lone \(x\) term), and \(e\) is the value after adding the square on both sides.

This process makes solving the quadratic equation manageable, as you'll see when we proceed to solving it.

Expanding Expressions

Expanding expressions involves opening up a compact mathematical expression into a lengthy one. When you see an expression like \(a(b + c)\), expanding it means applying the distributive property to remove the parentheses, transforming it into \(ab + ac\). This is a crucial step in many algebra problems, especially when dealing with equations that involve products of binomials or factoring.

In the quadratic equation that we have, the expression \(2x(2x-5) = 2\) involves expansion:

• Multiply \(2x\) by each term in the parenthesis: \(2x * 2x\) and \(2x * (-5)\), which results in \(4x^2 - 10x\).
• This transformation is essential because it converts the equation into a form that exposes each term, facilitating further algebraic manipulation, such as completing the square.
• After expanding "2x(2x - 5)", we move all terms to one side, yielding \(4x^2 - 10x - 2 = 0\).

This expanded form is what allows us to proceed to the next step of solving the equation by restructuring it.

Solving Equations

Once we've completed the square, solving the equation becomes a straightforward balancing act. We transform the quadratic into a perfect square trinomial which is then set equal to some number. This format makes it easy to solve using the square root property:

• Take the square root of both sides. Be sure to consider both positive and negative roots. This stems from the property that \((x - a)^2 = b\) implies \(x-a = \pm \sqrt{b}\).
• After deriving the principal square root, re-arrange the equation to isolate \(x\).
• For equations of the form \((x - p)^2 = q\), your next step is \(x - p = \pm \sqrt{q}\).
• Adjust the expression to solve for \(x\) by adding \(p\) on both sides, giving us the final solutions.
• In our specific problem, once the equation is \((x - \frac{5}{4})^2 = \frac{33}{16}\), we find \(x = \frac{5}{4} \pm \frac{\sqrt{33}}{4}\).

This method illustrates a reliable path for solving any quadratic equation when completed squares present us with intuitive solutions.