Problem 71 - Elementary Algebra

Question

Use the quadratic formula to solve the following. $$27 y(y+1)+2(3 y-2)=0$$

Step-by-Step Solution

Answer

The solutions are \( y = \frac{1}{9} \) and \( y = -\frac{4}{3} \).

Step 1: Expand and Simplify the Equation

First, distribute the terms in the equation: \[ 27y(y + 1) + 2(3y - 2) = 0 \]This gives:\[ 27y^2 + 27y + 6y - 4 = 0 \]Combine like terms:\[ 27y^2 + 33y - 4 = 0 \]

Step 2: Identify Coefficients for the Quadratic Formula

The quadratic equation is in the form \( ay^2 + by + c = 0 \) where:- \( a = 27 \)- \( b = 33 \)- \( c = -4 \)

Step 3: Apply the Quadratic Formula

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Substitute the values for \(a\), \(b\), and \(c\) into the formula:\[ y = \frac{-33 \pm \sqrt{33^2 - 4 \times 27 \times (-4)}}{2 \times 27} \]

Step 4: Calculate the Discriminant

Find the discriminant, which is \( b^2 - 4ac \):\[ 33^2 - 4 \times 27 \times (-4) = 1089 + 432 = 1521 \]

Step 5: Take the Square Root of the Discriminant

Take the square root of 1521:\[ \sqrt{1521} = 39 \]

Step 6: Solve for y Using the Plus or Minus in the Quadratic Formula

Plug the value back into the formula:\[ y = \frac{-33 \pm 39}{54} \]This yields two solutions:\[ y_1 = \frac{-33 + 39}{54} = \frac{6}{54} = \frac{1}{9} \]\[ y_2 = \frac{-33 - 39}{54} = \frac{-72}{54} = -\frac{4}{3} \]

Step 7: Write Down the Solutions

The solutions to the quadratic equation are \( y = \frac{1}{9} \) and \( y = -\frac{4}{3} \).

Key Concepts

Quadratic Equations

When dealing with quadratic equations, you are faced with expressions that take the form \( ay^2 + by + c = 0 \). These equations can have one unknown variable squared, hence the name quadratic. The main goal is to find values of the variable \( y \) that make the equation true. Quadratic equations can have:

• Two real distinct solutions
• One real repeated solution
• No real solutions

In our exercise, the equation \( 27y^2 + 33y - 4 = 0 \) is a classic quadratic form. Here, \( a = 27 \), \( b = 33 \), and \( c = -4 \). Solving it involves rearranging and simplifying to find the values of \( y \). You often use methods like factoring, completing the square, or applying the Quadratic Formula to tackle these problems effectively.

Discriminant

The discriminant is a key part of the quadratic formula, which helps to determine the nature of the roots of a quadratic equation. It is calculated using the expression \( b^2 - 4ac \). By assessing this value, you can know the kind of solutions you will get:

• If the discriminant is positive, there are two distinct real roots
• If it's zero, the equation has exactly one real root (a repeated root)
• If negative, the roots are complex or imaginary and not real

For our specific case, the discriminant was computed as follows:\[33^2 - 4 \times 27 \times (-4) = 1089 + 432 = 1521\]Since 1521 is a positive number, the quadratic equation \( 27y^2 + 33y - 4 = 0 \) has two distinct real solutions. This insight is crucial for deciding the method by which to solve the equation.

Algebraic Solutions

Finding algebraic solutions to quadratic equations typically involves using the quadratic formula, which is a powerful tool in algebra. The quadratic formula is:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Once you have the values for \( a \), \( b \), and \( c \) from your quadratic equation, you substitute these into the formula. In this exercise, we used:

• \( a = 27 \)
• \( b = 33 \)
• \( c = -4 \)

We calculated \( \pm \sqrt{1521} = 39 \), leading to:

• \( y_1 = \frac{-33 + 39}{54} = \frac{1}{9} \)
• \( y_2 = \frac{-33 - 39}{54} = -\frac{4}{3} \)

These are the solutions to the equation \( 27y^2 + 33y - 4 = 0 \). Employing the quadratic formula allows you to find values for \( y \) efficiently, which is particularly useful when the equation cannot be easily factored.