Problem 15
Question
solve the given quadratic equations, using the quadratic formula. Exercises \(5-8\) are the same as Exercises \(11-14\) of Section 7.2. $$3 y^{2}=3 y+2$$
Step-by-Step Solution
Verified Answer
The solutions for \( y \) are \( y = \frac{3 + \sqrt{33}}{6} \) and \( y = \frac{3 - \sqrt{33}}{6} \).
1Step 1: Rewriting Equation to Standard Form
Start by rewriting the given quadratic equation into the standard form, which is written as \( ax^2 + bx + c = 0 \). The given equation is \( 3y^2 = 3y + 2 \). Subtract \( 3y + 2 \) from both sides to get \( 3y^2 - 3y - 2 = 0 \).
2Step 2: Identifying a, b, and c
Identify the coefficients \( a \), \( b \), and \( c \) from the standard form equation \( 3y^2 - 3y - 2 = 0 \). Compare it to \( ax^2 + bx + c = 0 \) to see that \( a = 3 \), \( b = -3 \), and \( c = -2 \).
3Step 3: Using the Quadratic Formula
The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the values of \( a \), \( b \), and \( c \) into the formula: \( y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \).
4Step 4: Simplifying the Expression
Simplify inside the square root: \( (-3)^2 = 9 \) and \( 4 \cdot 3 \cdot (-2) = -24 \). Thus, the expression inside the square root becomes \( 9 - (-24) = 9 + 24 = 33 \).
5Step 5: Solving for y
The simplified quadratic formula is \( y = \frac{3 \pm \sqrt{33}}{6} \). Split this into the two possible solutions: \( y = \frac{3 + \sqrt{33}}{6} \) and \( y = \frac{3 - \sqrt{33}}{6} \). Both are simplified forms for \( y \).
Key Concepts
Quadratic FormulaStandard Form of a Quadratic EquationSolving EquationsMathematics Education
Quadratic Formula
When faced with a quadratic equation of the form \( ax^2 + bx + c = 0 \), the quadratic formula is a powerful tool to find its solutions. The formula is:
The term under the square root, \( b^2 - 4ac \), is called the discriminant. It helps determine the nature of the roots:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
The term under the square root, \( b^2 - 4ac \), is called the discriminant. It helps determine the nature of the roots:
- If the discriminant is positive, there are two distinct real solutions.
- If it equals zero, there is exactly one real solution.
- If it's negative, the solutions are complex or imaginary numbers.
Standard Form of a Quadratic Equation
Before applying the quadratic formula, it's essential to write the quadratic equation in its standard form: \( ax^2 + bx + c = 0 \). This form clearly presents the coefficients \( a \), \( b \), and \( c \), which are necessary to use the quadratic formula effectively.
Rewriting an equation to standard form involves moving all terms to one side of the equation, leaving zero on the other side.
For the given example, starting with \( 3y^2 = 3y + 2 \), you would subtract \( 3y + 2 \) from both sides: resulting in \( 3y^2 - 3y - 2 = 0 \).
This not only facilitates solving the equation but also enhances your understanding of how each component interacts in the equation.
Rewriting an equation to standard form involves moving all terms to one side of the equation, leaving zero on the other side.
For the given example, starting with \( 3y^2 = 3y + 2 \), you would subtract \( 3y + 2 \) from both sides: resulting in \( 3y^2 - 3y - 2 = 0 \).
This not only facilitates solving the equation but also enhances your understanding of how each component interacts in the equation.
Solving Equations
To solve a quadratic equation, especially using the quadratic formula, follow these steps:
- Rearrange the equation to standard form: \( ax^2 + bx + c = 0 \).
- Identify the coefficients \( a \), \( b \), and \( c \).* In our example, these are \( a = 3 \), \( b = -3 \), and \( c = -2 \).
- Substitute these values into the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
- Calculate the discriminant \( b^2 - 4ac \) to find the type and number of solutions.
- Simplify the expression, compute the final values, and write down the possible solutions.
Mathematics Education
Understanding and using quadratic equations is a fundamental skill in mathematics education. It lays the groundwork for exploring more complex algebraic concepts. Mastery of these equations aids in developing logical thinking and problem-solving skills.
In classroom settings, students often practice quadratic equations through exercises that enhance understanding and encourage exploration.
In classroom settings, students often practice quadratic equations through exercises that enhance understanding and encourage exploration.
- Solving quadratic equations improves critical thinking.
- It reinforces mathematical concepts like factoring and completing the square.
- Students gain confidence as they understand the application of quadratic equations in real-world problems.
Other exercises in this chapter
Problem 14
Solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$t^{2}+5 t-6=0$$
View solution Problem 15
Sketch the graph of each parabola by using the vertex, the \(y\) -intercept, and two other points, not including the \(x\) -intercepts. Check the graph using a
View solution Problem 15
In Exercises \(11-30,\) solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$n^{2}=4 n
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$$\text { Solve the given quadratic equations by factoring.}$$ $$R^{2}+12=7 R$$
View solution