Problem 43 - Algebra A Combined Function

Question

Solve each equation by completing the square. $$ y^{2}+y-7=0 $$

Step-by-Step Solution

Answer

The solutions are \( y = -\frac{1}{2} + \frac{\sqrt{29}}{2} \) and \( y = -\frac{1}{2} - \frac{\sqrt{29}}{2} \).

Step 1: Move Constant to Other Side

Start by moving the constant term \( -7 \) to the right side of the equation. This gives us: \(y^2 + y = 7\\).

Step 2: Find the Perfect Square

To complete the square, we need to make the left side of the equation a perfect square trinomial. Take the coefficient of the linear term \( y \), which is \( 1 \), halve it to get \(\frac{1}{2}\\), and then square it to get \( \frac{1}{4}\). Add and subtract \( \frac{1}{4}\) inside the equation: \(y^2 + y + \frac{1}{4} - \frac{1}{4} = 7\\).

Step 3: Write the Equation as a Perfect Square Trinomial

Now, the left side of equation is: \(y^2 + y + \frac{1}{4} = \left(y + \frac{1}{2}\right)^2.\) Thus, the equation becomes: \(\left(y + \frac{1}{2}\right)^2 = 7 + \frac{1}{4}\\).

Step 4: Simplify the Right Side

Combine the terms on the right side of the equation \(7 + \frac{1}{4}\) by getting a common denominator: \(7 = \frac{28}{4}\). Thus, \(\frac{28}{4} + \frac{1}{4} = \frac{29}{4}.\) The equation becomes: \(\left(y + \frac{1}{2}\right)^2 = \frac{29}{4}\\).

Step 5: Solve for y

Take the square root of both sides to find the value of \( y \): \(y + \frac{1}{2} = \pm \frac{\sqrt{29}}{2}\\). Solve for \( y \) by subtracting \( \frac{1}{2} \): \( y = -\frac{1}{2} \pm \frac{\sqrt{29}}{2}\\).

Step 6: Find the Solution

This gives the two solutions: \( y = -\frac{1}{2} + \frac{\sqrt{29}}{2} \) and \( y = -\frac{1}{2} - \frac{\sqrt{29}}{2}\).

Key Concepts

Quadratic equations

Quadratic equations are mathematical expressions that form a curve called a parabola when plotted on a graph. These equations typically take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the unknown variable.

The power or degree of the equation is 2, which means the highest exponent of the variable \(x\) is 2. This characteristic shapes the quadratic format.

• The constant \(a\) determines the direction and the width of the parabola, with positive \(a\) values opening upward and negative ones opening downward.
• The constant \(b\) influences the location of the vertex or the highest/lowest point of the parabola along the axis.
• Lastly, \(c\) provides the point at which the parabola intercepts the y-axis.

Quadratic equations can be solved using different methods, like factoring, using the quadratic formula, or by completing the square, as we did in the given problem.

Solving equations

Solving equations, particularly quadratic ones, involves finding the values of the variable that make the equation true. In quadratic equations, this typically results in two solutions because of the squaring of the variable.

In the problem provided, we aimed to find the solutions of \(y^2 + y - 7 = 0\) by completing the square. Here’s a concise review of the steps involved:

• First, the equation is rearranged to isolate the square or variable terms from the constant, \(y^2 + y = 7\).
• Next, we complete the square, which involves creating a perfect square trinomial out of the expression \(y^2 + y\).
• We add the necessary value to both sides of the equation to maintain equality after forming a perfect square.
• After simplifying both sides, we take the square root to find the values of \(y\).
• Finally, we solve for \(y\) to get the two solutions.

By applying these steps, we derived the solutions as: \( y = -\frac{1}{2} + \frac{\sqrt{29}}{2} \) and \( y = -\frac{1}{2} - \frac{\sqrt{29}}{2}\). These solutions satisfy the original equation when substituted back.

Perfect square trinomial

A perfect square trinomial is a special type of algebraic expression and an essential concept in algebra. It is an expression that can be factored into a square of a binomial. This means it looks like \((x + a)^2\), which expands to \(x^2 + 2ax + a^2\).

In the problem, the process of completing the square requires manipulating the expression into a perfect square trinomial. Here's how it works and why it's important:

• Completing the square turns part of the quadratic expression \(y^2 + y\) into a form that can be easily factored.
• We add \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \) to both sides of the equation to form a trinomial \(y^2 + y + \frac{1}{4}\), which equals \((y + \frac{1}{2})^2\).
• This operation doesn’t change the equation’s value, so we add \( \frac{1}{4} \) to ensure fairness while transforming the equation.

The mastery of identifying and using perfect square trinomials is crucial for solving many algebraic problems, making this technique a powerful tool in a student's mathematical toolkit.