Problem 65 - Elementary Algebra

Question

For the following problems, solve the equations by completing the square or by using the quadratic formula. $$ 5 a^{2}+2 a-6=0 $$

Step-by-Step Solution

Answer

Based on the given problem, we were asked to solve the quadratic equation $5a^2 + 2a - 6 = 0$. We first solved it using the method of completing the square, and then by using the quadratic formula. In both cases, we found that the solutions to the given equation are $a = -\frac{1}{5} +\sqrt{\frac{31}{25}}$ or $a = -\frac{1}{5} -\sqrt{\frac{31}{25}}$.

Step 1: Solve by Completing the Square

First, we'll solve the equation by completing the square: $$ 5a^2 + 2a - 6 = 0 $$ Start by making sure the equation is set to zero and factor out any common factor from the quadratic terms (in this case, there's no common factor).

Step 2: Write the Equation in Completed Square Form

Now, let's complete the square for the expression inside the parentheses: 1. Add and subtract the square of half of the linear coefficient to the equation. $$ 5a^2 + 2a + (\frac{2}{2*5})^2 - (\frac{2}{2*5})^2 - 6 = 0 $$ Calculating the values gives: $$ 5a^2 + 2a + \frac{1}{25} - \frac{1}{25} - 6 = 0 $$ 2. Factor the perfect square trinomial and simplify the result. $$ 5(a + \frac{1}{5})^2 - \frac{1}{5} - 6 = 0 $$

Step 3: Solve the Equation after Completing the Square

Now, solve the equation by applying the square root property: 1. Isolate the square term. $$ 5(a + \frac{1}{5})^2 = \frac{1}{5} + 6 $$ 2. Simplify and eliminate the constant factor. $$ (a + \frac{1}{5})^2 = \frac{31}{25} $$ 3. Apply square root property and solve for a. $$ a + \frac{1}{5} = \pm\sqrt{\frac{31}{25}} $$ $$ a = -\frac{1}{5} \pm\sqrt{\frac{31}{25}} $$ So the solutions to the equation are: $$ a = -\frac{1}{5} +\sqrt{\frac{31}{25}} \quad\text{or}\quad a = -\frac{1}{5} -\sqrt{\frac{31}{25}} $$

Step 4: Solve Using Quadratic Formula

We can also use the quadratic formula to solve this equation. The quadratic formula is given by: $$ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ In this case, the coefficients are as follows: - a = 5 - b = 2 - c = -6 Now, plug the coefficients into the quadratic formula and find the solutions. $$ a = \frac{-(2) \pm \sqrt{(2)^2 - 4(5)(-6)}}{2(5)} $$ Simplifying the expression gives: $$ a = \frac{-2 \pm \sqrt{124}}{10} $$ Factoring out a 2 from the numerator, we get: $$ a = \frac{-1 \pm \sqrt{\frac{31}{5}}}{5} $$ Thus, the solutions are: $$ a = -\frac{1}{5} +\sqrt{\frac{31}{25}} \quad\text{or}\quad a = -\frac{1}{5} -\sqrt{\frac{31}{25}} $$ This is the same result we got using the completing the square method.

Key Concepts

Completing the Square

Completing the square is a technique used to solve quadratic equations. It helps in transforming a quadratic equation into a perfect square trinomial, which is easier to solve. The general formula of a quadratic equation is given by \( ax^2 + bx + c = 0 \). If necessary, begin by ensuring your leading coefficient (the coefficient of \( x^2 \)) is 1. If not, you should divide the entire equation by \( a \).

• First, move the constant term to the other side of the equation. For example, if starting with \( ax^2 + bx = -c \).
• Calculate \( \left( \frac{b}{2a} \right)^2 \) and add this square to both sides. This step creates a perfect square trinomial on one side.
• Now the equation takes the form: \((x + d)^2 = e \), where \( d \) is half the linear coefficient, and \( e \) is the adjusted constant.
• Finally, solve for \( x \) by taking the square root of both sides, bearing in mind to consider both the positive and negative square roots.

In our specific example, after completing these steps, handy factoring, and simplifying, the result led us to find the solutions for \( a \).

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation \( ax^2 + bx + c = 0\). It's essentially a one-stop solution method that can handle equations where other techniques might be cumbersome. The formula itself is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]The terms \( b^2 - 4ac \) are called the discriminant. They help determine the nature of the roots:

• If the discriminant is \(> 0\), there are two distinct real roots.
• If the discriminant is \(= 0\), there is exactly one real root (a repeated root).
• If the discriminant is \(< 0\), the roots are complex.

In the example, we substituted \( a = 5 \), \( b = 2 \), and \( c = -6 \) into the formula. Simplifying the expression confirmed the solutions obtained by completing the square, showcasing the consistency between methods.

Algebraic Solutions

Using algebraic methods to solve quadratic equations involves leveraging algebra rules for manipulating equations. Both completing the square and the quadratic formula are algebraic solutions and serve well when tackling quadratic problems. They allow us to find exact numeric or algebraic values of \( x \) (or any variable, like \( a \) in the exercise), ensuring we can express the solutions in a precise mathematical form.
Each method comes with its own set of steps and procedures:

• Completing the Square: Good for understanding the roots' relation and when rearranging terms results in a neat perfect square.
• Quadratic Formula: A reliable fallback when factorization isn't straightforward. It uses direct formula application.

Both methods ultimately guide us to the same solutions, which proves their effectiveness and reliability in solving quadratics.