Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ \frac{5}{3} x^{2}-x=\frac{1}{3} $$

Short answer

The real solutions are \( x = \frac{3 + \sqrt{29}}{10} \) and \( x = \frac{3 - \sqrt{29}}{10} \).

Step-by-step solution

- Bring Equation to Standard Form

Start by bringing the given equation \( \frac{5}{3} x^{2} - x = \frac{1}{3} \) to the standard quadratic form \( ax^2 + bx + c = 0 \). Subtract \( \frac{1}{3} \) from both sides to get: \[ \frac{5}{3} x^{2} - x - \frac{1}{3} = 0 \]

- Identify Coefficients

Identify the coefficients \( a \), \( b \), and \( c \) from the equation \( \frac{5}{3} x^2 - x - \frac{1}{3} = 0 \). Here, \( a = \frac{5}{3} \), \( b = -1 \), and \( c = -\frac{1}{3} \).

- Write Down the Quadratic Formula

Recall the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]

- Substitute the Coefficients into the Formula

Substitute \( a = \frac{5}{3} \), \( b = -1 \), and \( c = -\frac{1}{3} \) into the quadratic formula: \[ x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4 \cdot \frac{5}{3} \cdot -\frac{1}{3}}}}}{2 \cdot \frac{5}{3}} \]

- Simplify the Equation Under the Square Root

Simplify under the square root: \[ x = \frac{{1 \pm \sqrt{{1 - 4 \cdot \frac{5}{3} \cdot -\frac{1}{3}}}}}{2 \cdot \frac{5}{3}} = \frac{{1 \pm \sqrt{{1 + \frac{20}{9}}}}}{\frac{10}{3}} \]

- Further Simplify the Expression

Continue simplifying: \[ x = \frac{{1 \pm \sqrt{{1 + \frac{20}{9}}}}}{\frac{10}{3}} = \frac{{1 \pm \sqrt{{\frac{9}{9} + \frac{20}{9}}}}}{\frac{10}{3}} = \frac{{1 \pm \sqrt{{\frac{29}{9}}}}}{\frac{10}{3}} \]

- Simplify the Square Root

Take the square root of the fraction: \[ x = \frac{{1 \pm \sqrt{\frac{29}{9}}}}{\frac{10}{3}} = \frac{{1 \pm \frac{\sqrt{29}}{3}}}{\frac{10}{3}} \]

- Simplify the Final Expression

Divide the numerator by the denominator: \[ x = \frac{1 \pm \frac{\sqrt{29}}{3}}{\frac{10}{3}} = \frac{3(1 \pm \frac{\sqrt{29}}{3})}{10} = \frac{3 \pm \sqrt{29}}{10} \]

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool to solve quadratic equations of the form \(ax^2 + bx + c = 0\). It provides the solutions directly, even when factoring seems difficult or impossible.
To recall, the quadratic formula is:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
The formula calculates the roots (solutions) of the equation by substituting the coefficients \(a\), \(b\), and \(c\).
Let's break down the formula further:

• \(-b\) means you take the opposite sign of the coefficient \(b\).
• \(\sqrt{b^2 - 4ac}\) is called the discriminant. It determines the nature of the roots.
• \(2a\) is the denominator that scales the numerator correctly.

Using the quadratic formula ensures we consider all potential roots and provides a systemic way to handle quadratic equations across diverse problems.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols.
In context, solving a quadratic equation involves several algebraic techniques, such as:

• Rearranging terms: Ensure the equation is in the standard form \(ax^2 + bx + c = 0\).
• Identifying coefficients: Determine the values of \(a\), \(b\), and \(c\) from the equation.
• Substituting values: Plug these coefficients into the quadratic formula.
• Performing operations: Carry out arithmetic operations like addition, multiplication, and square root extraction to simplify expressions.

For example, in the equation \( \frac{5}{3} x^2 - x = \frac{1}{3} \), we first bring all terms to one side:
\(\frac{5}{3} x^2 - x - \frac{1}{3} = 0 \).
While dealing with complex transformations, maintaining balance and systematically applying algebraic rules are key for correctly solving the equation.

Real Solutions

When solving quadratic equations using the quadratic formula, the discriminant \(b^2 - 4ac\) plays a crucial role in determining the nature of the solutions.
Here's how to understand real solutions:

• If \(b^2 - 4ac > 0\), the equation has two distinct real solutions.
• If \(b^2 - 4ac = 0\), the equation has exactly one real solution, also referred to as a repeated or double root.
• If \(b^2 - 4ac < 0\), the equation has no real solutions, which means the roots are complex numbers.

In our example, after substituting the coefficients into the formula:
\( x = \frac{1 \pm \sqrt{ \frac{29}{9} }}{ \frac{10}{3} } \), we see the discriminant is positive \(29/9\).
Thus, \(\frac{29}{9}\) is greater than zero, confirming the quadratic equation has two distinct real solutions.
The final simplified solutions are \( x = \frac{3 + \sqrt{29}}{10} \) and \( x = \frac{3 - \sqrt{29}}{10} \).
Always check the discriminant first to understand the nature of the solutions you're working with.