Problem 40 - Precalculus Student Solutions Manual 5th

Question

Solve each equation by completing the square. $$4 x^{2}-3 x-10=0$$

Step-by-Step Solution

Answer

The solutions are x = 2 or x = -\(\frac{5}{4}\).

Step 1: Move the constant term to the other side

To start with, move the constant term to the right side of the equation. 4x^2 - 3x - 10 = 0 Add 10 to both sides: 4x^2 - 3x = 10

Step 2: Divide by the leading coefficient

Divide all terms by the leading coefficient, which is 4, to make the coefficient of x^2 equal to 1: \(\frac{4x^2}{4} - \frac{3x}{4} = \frac{10}{4}\)Simplify: x^2 - \(\frac{3}{4}x\) = \(\frac{5}{2}\)

Step 3: Complete the square

To complete the square, take half of the coefficient of x (which is \(\frac{3}{4}\)), square it, and add it to both sides of the equation: \[\left( \frac{-3}{4} \times \frac{1}{2} \right)^2 = \left( \frac{3}{8} \right)^2 = \frac{9}{64}\]Add \(\frac{9}{64}\) to both sides:x^2 - \(\frac{3}{4}x\) + \(\frac{9}{64}\) = \(\frac{5}{2} + \frac{9}{64}\)

Step 4: Simplify the right side

Combine the terms on the right side:\(\frac{5}{2}\) is the same as \(\frac{160}{64}\), so:\(\frac{160}{64} + \frac{9}{64} = \frac{169}{64}\)So, the equation becomes: \[x^2 - \frac{3}{4}x + \frac{9}{64} = \frac{169}{64}\]

Step 5: Write the left side as a squared binomial

The left side is a perfect square trinomial. Write it as a squared binomial:\[\left(x - \frac{3}{8}\right)^2 = \frac{169}{64}\]

Step 6: Solve for x

Solve for x by taking the square root of both sides:\[x - \frac{3}{8} = \pm \sqrt{\frac{169}{64}}\]Since the square root of \(\frac{169}{64}\) is \(\frac{13}{8}\), we get:\[x - \frac{3}{8} = \pm \frac{13}{8}\]Now, solve for x:\[x = \frac{3}{8} + \frac{13}{8} = \frac{16}{8} = 2\]or\[x = \frac{3}{8} - \frac{13}{8} = \frac{-10}{8} = -\frac{5}{4}\]

Key Concepts

quadratic equations

Quadratic equations are polynomial equations of degree 2. They have the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable. The highest exponent of \(x\) in a quadratic equation is 2. This type of equation can have up to two solutions.

To solve a quadratic equation, you can use various methods like factoring, using the quadratic formula, or completing the square—which is demonstrated in the exercise solution above.

solving equations

Solving equations involves finding the values of the variable that make the equation true. For quadratic equations, solving can be more complex due to the squared term. In the exercise provided, we used the 'completing the square' method. Here are the general steps:

1. Rearrange the equation to isolate the constant term on one side.

2. Divide all terms by the leading coefficient to make the coefficient of \(x^2\) equal to 1.

3. Complete the square by adding and subtracting the square of half the coefficient of \(x\).

4. Write the left side of the equation as a squared binomial.

5. Solve for \(x\) by taking the square root of both sides.

algebra techniques

Algebra involves various techniques to manipulate and solve equations. Let's focus on three key techniques used for solving quadratic equations:

- **Isolating the variable:** This step involves rearranging the equation so that the variable you want to solve for is on one side of the equation.

- **Completing the square:** Completing the square transforms a quadratic equation into a perfect square form \((x - p)^2 = q\), making it easier to solve.

- **Taking square roots:** Once a quadratic equation is in the form of a perfect square, you can take the square root of both sides to find the values of the variable.