Problem 57 - College Algebra

Question

Solve each equation in Exercises \(47-64\) by completing the square. $$x^{2}+3 x-1=0$$

Step-by-Step Solution

Answer

The roots of the quadratic equation are \( x = - \frac{3}{2} + \sqrt{1+\left(\frac{3}{2}\right)^{2}} \) and \( x = - \frac{3}{2} - \sqrt{1+\left(\frac{3}{2}\right)^{2}} \)

Step 1: Group the quadratic and linear terms

Rewrite the given quadratic equation \(x^{2}+3 x-1=0\) in the form \(x^{2} + 3x = 1\) by moving the constant term on the other side of the equation.

Step 2: Find the value to complete the square

Rearrange the equation: \(x^{2}+3 x + \left(\frac{3}{2}\right)^{2} - \left(\frac{3}{2}\right)^{2} = 1\). This has been done in order to put the left side of the equation into the standard form of a perfect square \((x-a)^{2}\). Here, \(\left(\frac{3}{2}\right)^{2}\) is the square of half the value of x's coefficient in the term \(3x\)

Step 3: Put the left hand side into perfect square form

Rewrite the equation to get it into a perfect square form: \(\left(x+\frac{3}{2}\right)^{2} - \left(\frac{3}{2}\right)^{2} = 1\). Here, \(\left(x+\frac{3}{2}\right)^{2}\) is a perfect square

Step 4: Solve for x

Simplify the equation: \( \left(x+\frac{3}{2}\right) = \pm \sqrt{1+\left(\frac{3}{2}\right)^{2}} \). Then, solve for x to get the roots: \( x = - \frac{3}{2} \pm \sqrt{1+\left(\frac{3}{2}\right)^{2}} \). This gives the two roots of the given quadratic equation.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation characterized by the presence of a variable raised to the second power. They generally take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).
Quadratics are everywhere around us, from physics to finance. Understanding how to solve them can open doors to various exciting fields. Some of the essential components of a quadratic equation are:

• Quadratic Term: This is the \(ax^2\) term, indicating the parabola's shape.
• Linear Term: The \(bx\) term contributes to the direction and position of the parabola.
• Constant Term: The \(c\) provides the y-intercept of the curve on a graph.

By solving quadratic equations, whether by factoring, using the quadratic formula or by completing the square, we find the points where the curve crosses the x-axis. These points are called "roots" or "solutions."

Perfect Square Form

The perfect square form is a mathematical technique used to simplify the process of solving quadratic equations. By transforming a quadratic expression into a perfect square trinomial, we make it much easier to solve.
A perfect square trinomial is one that can be expressed as the square of a binomial. For instance, \((x+a)^2 = x^2 + 2ax + a^2\) is a perfect square. In our exercise, the original equation \(x^2 + 3x - 1 = 0\) was transformed by following these steps:

• Move the constant to the other side, resulting in \(x^2 + 3x = 1\).
• Find the number that completes the square by taking half of the \(x\)-coefficient, squaring it, and adding it inside the equation.
• Transform the equation into its perfect square form, \(\left(x + \frac{3}{2}\right)^2\).

This form allows us to easily apply square root operations to find the solutions. By working the equation into a perfect square, you set the stage for clear and manageable solutions.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. In our exercise, this meant finding the values of \(x\) that satisfy \(x^2 + 3x - 1 = 0\) after transforming the equation using completing the square.
Once the equation is written in perfect square form, you can solve it by taking the square root of both sides. This step leads to two potential solutions because squaring operations have this property of reversing for both positive and negative roots. Follow these steps:

• Write the perfect square equation: \(\left(x + \frac{3}{2}\right)^2 = 1 + \left(\frac{3}{2}\right)^2\).
• Take the square root of both sides, leading to \(x + \frac{3}{2} = \pm \sqrt{\text{right side}}\).
• Solve for \(x\) by isolating the variable, which produces two solutions because of the \(\pm\) sign.

Presenting equations in this way not only provides a straightforward path to solving but also gives insight into the nature of the solutions.