Problem 16 - Elementary Algebra

Question

For the following problems, solve the equations by completing the square. $$ b^{2}+5 b-3=0 $$

Step-by-Step Solution

Answer

Question: Solve the quadratic equation \(b^2 + 5b - 3 = 0\) by completing the square. Answer: \(b = -\frac{5}{2} \pm \sqrt{\frac{37}{4}}\)

Step 1: Set up the quadratic equation correctly

The given equation is: $$ b^2 + 5b -3 = 0 $$ To complete the square, we first need to move the constant term, -3, to the other side of the equation. So, we get: $$ b^2 + 5b = 3 $$

Step 2: Add the square of half the coefficient of the linear term to both sides

The coefficient of the linear term is 5. Half of the coefficient of the linear term is \(\frac{5}{2}\), and the square of that is \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\). So, let's add this to both sides of the equation: $$ b^2 + 5b + \frac{25}{4} = 3 + \frac{25}{4} $$

Step 3: Rewrite the left side as a square of a binomial and simplify the right side

The left side can now be factored as a square of a binomial, and the right side simplified: $$ \left(b + \frac{5}{2}\right)^2 = \frac{37}{4} $$

Step 4: Take the square root of both sides

We can now take the square root of both sides of the equation: $$ b + \frac{5}{2} = \pm \sqrt{\frac{37}{4}} $$

Step 5: Isolate the variable b

Finally, we need to isolate the variable b by subtracting \(\frac{5}{2}\) from both sides: $$ b = -\frac{5}{2} \pm \sqrt{\frac{37}{4}} $$ So, our final answer is: $$ b = -\frac{5}{2} \pm \sqrt{\frac{37}{4}} $$

Key Concepts

Solve Quadratic Equations

Quadratic equations are polynomial equations of the second degree, typically written in the form \( ax^2 + bx + c = 0 \). To solve these equations, we aim to find the values of the variable \( x \) that make the equation true. There are various methods to solve quadratic equations, such as factoring, using the quadratic formula, graphing, or completing the square.

Completing the square is a powerful technique, especially when the quadratic equation does not factor easily. This method involves rearranging the equation to form a perfect square trinomial on one side, allowing us to then take the square root of both sides to solve for the variable. By mastering this method, students gain a greater understanding of the structure of quadratic equations and become more adept at manipulating algebraic expressions.

Square of a Binomial

The square of a binomial is a fundamental algebraic identity that expresses the product of a binomial multiplied by itself. If we have any binomial of the form \( (a + b) \), its square is \( (a + b)^2 = a^2 + 2ab + b^2 \). This is a pattern that students often memorize.

When completing the square in the context of solving quadratic equations, we strategically add and subtract terms to create a perfect square trinomial, which can then be written as the square of a binomial. The importance of recognizing this pattern cannot be overstated, as it not only simplifies the process of solving quadratics but also builds a strong foundation for more advanced algebraic concepts. In our example, we transformed \( b^2 + 5b \) into \( (b + \frac{5}{2})^2 \), demonstrating the practical application of this concept.

Isolate the Variable

Isolating the variable is the final and crucial step in solving equations. It involves manipulating the equation so that the variable of interest is alone on one side of the equation. This allows us to identify its value or values clearly.

In the context of completing the square, after we have expressed one side of the equation as the square of a binomial, we proceed by taking the square root of both sides. This helps us get rid of the square and bring us one step closer to isolating the variable. For example, after taking the square root, we end up with \( b + \frac{5}{2} = \pm \sqrt{\frac{37}{4}} \). The final step to isolate \( b \) is to subtract \( \frac{5}{2} \) from both sides, making the solution explicit. Understanding this step is essential because it forms the basis for solving not only quadratics but virtually any algebraic equation.