Problem 60 - Precalculus Student Solutions Manual 5th

Question

Solve each equation using the quadratic formula. $$\frac{2}{3} x^{2}+\frac{1}{4} x=3$$

Step-by-Step Solution

Answer

The solutions are \[x = \frac{-3 + 3 \sqrt{129}}{16}\] and \[x = \frac{-3 - 3 \sqrt{129}}{16}\].

Step 1: Rewrite the equation in standard form

The standard form of a quadratic equation is given by \[ax^2 + bx + c = 0\] First, move all terms to one side of the equation to set it to zero: \[\frac{2}{3} x^2 + \frac{1}{4} x - 3 = 0\]

Step 2: Identify coefficients

Compare the given equation with \[ax^2 + bx + c = 0\] to identify the coefficients: \[a = \frac{2}{3}, \, b = \frac{1}{4}, \, c = -3\]

Step 3: Write down the quadratic formula

The quadratic formula to solve \[ax^2 + bx + c = 0\] is: \[x = \frac{-b \, \text{±} \, \sqrt{b^2 - 4ac}}{2a}\]

Step 4: Substitute the identified coefficients into the formula

Substitute \[a = \frac{2}{3}, \, b = \frac{1}{4}, \, c = -3\] into the quadratic formula: \[x = \frac{-\frac{1}{4} \, \text{±} \, \sqrt{ \big( \frac{1}{4} \big)^2 - 4 \big( \frac{2}{3} \big) \big( -3 \big) }}{2 \big( \frac{2}{3} \big)}\]

Step 5: Simplify under the square root

Calculate the term under the square root (discriminant) \[(b^2 - 4ac)\]: \[ \big( \frac{1}{4} \big)^2 - 4 \big( \frac{2}{3} \big) (-3) = \frac{1}{16} + 8 = \frac{1}{16} + \frac{128}{16} = \frac{129}{16}\]

Step 6: Simplify the quadratic formula expression

Substitute back into the formula: \[x = \frac{ -\frac{1}{4} \, \text{±} \, \sqrt{ \frac{129}{16}}}{ \frac{4}{3} }\] Simplify under the square root and solve: \[x = \frac{ -\frac{1}{4} \, \text{±} \, \frac{ \sqrt{129}}{4}}{ \frac{4}{3} } = \frac{ \frac{-1 \, \text{±} \, \sqrt{129}}{4} }{ \frac{4}{3} }\] Continue simplification: \[x = \frac{ -1 \, \text{±} \, \sqrt{129} }{4} \times \frac{3}{4} = \frac{ -3 \, \text{±} \, 3 \sqrt{129}}{16} = \frac{-3}{16} \, \text{±} \, \frac{3 \sqrt{129}}{16}\]

Step 7: Find the two solutions

The final solutions are: \[x = \frac{-3 + 3 \sqrt{129}}{16}\] and \[x = \frac{-3 - 3 \sqrt{129}}{16}\]

Key Concepts

Standard Form of a Quadratic Equation

A quadratic equation is one that can be written in the form \( ax^2 + bx + c = 0 \).Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). This is known as the standard form of a quadratic equation. Understanding this form is crucial because it allows us to use various mathematical techniques, such as factoring or the quadratic formula, to find the solutions for \(x\).

Discriminant

The discriminant is a key part of the quadratic formula located under the square root, represented by \( b^2 - 4ac \). The value of the discriminant determines the nature of the roots of a quadratic equation:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real root (or a repeated root).
• If \( b^2 - 4ac < 0 \), there are no real roots; instead, the solutions are complex numbers.

Knowing the value of the discriminant before solving can provide insight into the type of solutions to expect.

Solving Quadratic Equations

Solving quadratic equations often involves using the quadratic formula, especially when the equation does not factor easily. The quadratic formula is given by:\(x = \frac{-b \text{±} \sqrt{b^2 - 4ac}}{2a}\)To solve the given equation \( \frac{2}{3} x^2 + \frac{1}{4} x - 3 = 0 \), we first identify the coefficients \( a \), \( b \), and \( c \) as follows:

• \(a = \frac{2}{3}\)
• \(b = \frac{1}{4}\)
• \(c = -3\)

Substitute these values into the quadratic formula and simplify step-by-step to find the roots of the equation. After simplification, you get the solutions:\( x = \frac{-3 + 3 \sqrt{129}}{16} \) and \( x = \frac{-3 - 3 \sqrt{129}}{16} \).These solutions give the values of \(x\) that satisfy the original quadratic equation.