Problem 80 - Elementary Algebra

Question

Solve using the quadratic formula. $$ 5 t(t-1)=t-4 $$

Step-by-Step Solution

Answer

There are no real solutions.

Step 1: Rewrite the Equation in Standard Form

First, expand the left side of the equation: \[5t(t-1) = 5t^2 - 5t\]Now, subtract \(t\) and \(4\) from both sides to set the equation to zero:\[5t^2 - 5t - t + 4 = 0\]This simplifies to the standard form:\[5t^2 - 6t + 4 = 0\]

Step 2: Identify Coefficients

In the quadratic equation \(ax^2 + bx + c = 0\), identify the coefficients:\[a = 5, \quad b = -6, \quad c = 4\]

Step 3: Apply the Quadratic Formula

Use the quadratic formula:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plug in the identified coefficients:\[ t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4\cdot5\cdot4}}{2\cdot5} \]

Step 4: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\):\[(-6)^2 - 4\cdot5\cdot4 = 36 - 80 = -44\]Since the discriminant is negative, there are no real solutions.

Key Concepts

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is an essential concept. It lays the groundwork for solving quadratic equations using various methods, including the quadratic formula. A quadratic equation typically appears as: \[ax^2 + bx + c = 0\] where:

• \(a\), called the leading coefficient, is the coefficient of the squared term \(x^2\).
• \(b\) is the coefficient of the linear term \(x\).
• \(c\) is the constant term.

To use the quadratic formula, ensure the equation is in this standard form. This often involves expanding brackets, simplifying expressions, and moving terms around to set the equation to equal zero. In our example, the equation was reformatted from \(5t(t-1)=t-4\) to \(5t^2 - 6t + 4 = 0\). This step is crucial as it allows us to correctly identify the coefficients \(a\), \(b\), and \(c\), which are necessary to apply the quadratic formula effectively.

Discriminant

The discriminant is a powerful tool that predicts the nature of the solutions to a quadratic equation. It is part of the quadratic formula and is found within the square root:\[\Delta = b^2 - 4ac\] This small expression can tell you a lot about the solutions:

• If \(\Delta > 0\), the quadratic equation has two distinct real solutions.
• If \(\Delta = 0\), there is exactly one real solution, referred to as a repeated or double root.
• If \(\Delta < 0\), no real solutions exist, and the solutions are complex.

In the given problem, the discriminant was calculated as \((-6)^2 - 4 \, \cdot \, 5 \, \cdot \, 4 = 36 - 80 = -44\). Since \(\Delta = -44\) is negative, this indicates that the quadratic equation has no real solutions and will instead have complex solutions.

Complex Solutions

When a quadratic equation has a negative discriminant, it results in complex solutions rather than real ones. Complex numbers include an imaginary part, usually indicated by the symbol \(i\), where \(i\) is the square root of \(-1\). The solutions take the form:\[x = \frac{-b \pm \sqrt{\Delta}}{2a}\] where \(\Delta = b^2 - 4ac\). Since \(\Delta\) is negative, the square root of a negative number results in an imaginary component. For our equation, \(\Delta = -44\), the solutions would look like:\[t = \frac{-(-6) \pm \sqrt{-44}}{2 \cdot 5}\] Simplifying further, we'd calculate:

• The real part: \(\frac{6}{10} = 0.6\)
• The imaginary part from \(\sqrt{-44} = i\sqrt{44}\)

Thus, the complex solutions of the quadratic equation are:\[t = 0.6 \pm i\sqrt{11}\] Understanding complex solutions broadens your mathematical toolkit, enabling you to solve equations even when they do not have real roots.