Problem 79 - Elementary Algebra

Question

Solve using the quadratic formula. $$ 3 t(t-2)+4=0 $$

Step-by-Step Solution

Answer

The solutions are complex: \( t = 1 \pm \frac{i\sqrt{3}}{3} \).

Step 1: Expand the Quadratic Expression

First, we need to expand the given quadratic expression. The expression is \( 3t(t-2)+4=0 \). Expand it to get: \[3t^2 - 6t + 4 = 0.\]

Step 2: Identify Coefficients for the Quadratic Formula

To use the quadratic formula, we need to identify the coefficients \(a\), \(b\), and \(c\) from the standard form \( ax^2 + bx + c = 0 \). Compare with \( 3t^2 - 6t + 4 = 0 \), we get \( a = 3 \), \( b = -6 \), and \( c = 4 \).

Step 3: Write the Quadratic Formula

The quadratic formula is given by: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] Let's substitute the values for \( a \), \( b \), and \( c \).

Step 4: Calculate the Discriminant

The discriminant \( b^2 - 4ac \) determines the nature of the roots. Calculate: \[(-6)^2 - 4 \times 3 \times 4 = 36 - 48 = -12.\] Since the discriminant is negative, there are no real roots.

Step 5: Simplify Using the Quadratic Formula

Since the discriminant is negative, solve for the imaginary roots: \[t = \frac{-(-6) \pm \sqrt{-12}}{2 \times 3} = \frac{6 \pm \sqrt{-12}}{6} = \frac{6 \pm 2i\sqrt{3}}{6} = 1 \pm \frac{i\sqrt{3}}{3}.\]

Step 6: Finalize the Solution

The solutions to the quadratic equation \(3t^2 - 6t + 4 = 0 \) are: \[ t = 1 + \frac{i\sqrt{3}}{3} \quad \text{and} \quad t = 1 - \frac{i\sqrt{3}}{3}. \]

Key Concepts

Coefficients

When dealing with quadratic equations, understanding the coefficients is crucial. Coefficients are the numbers in front of the variables in an equation. In the quadratic equation, the standard form is expressed as \( ax^2 + bx + c = 0 \). Here, the letters \( a \), \( b \), and \( c \) represent the coefficients:

• \( a \) is the coefficient of \( x^2 \).
• \( b \) is the coefficient of \( x \).
• \( c \) is the constant term.

In the exercise given, we first expanded the expression to get \( 3t^2 - 6t + 4 = 0 \). From this, we identified the coefficients as \( a = 3 \), \( b = -6 \), and \( c = 4 \). These coefficients are essential for plugging into the quadratic formula, which helps in finding the solutions to the equation. Realizing their roles in shaping the equation and guiding the next steps is a fundamental skill in algebra.

Discriminant

The discriminant is a key component in the quadratic formula, and it gives insight into the nature of the roots of a quadratic equation. It is found by calculating \( b^2 - 4ac \). Here’s what it means:

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it's zero, there is exactly one real root (which is also called a double root).
• If the discriminant is negative, the roots are not real; they are complex or imaginary.

In our example, we calculated the discriminant to be \( (-6)^2 - 4 \times 3 \times 4 = 36 - 48 = -12 \). Since the value is negative, it tells us there are no real roots for the equation \( 3t^2 - 6t + 4 = 0 \). Instead, the roots are imaginary, requiring further calculation for complex solutions.

Imaginary Roots

In some cases, when the discriminant is negative, the roots of the quadratic equation are imaginary. Imaginary numbers are used to describe these solutions. The imaginary unit is denoted by \( i \), where \( i = \sqrt{-1} \). The quadratic formula allows for calculating these roots even when the value under the square root is negative. The roots are expressed in the form of complex numbers:

• The real part: given by the expression \( \frac{-b}{2a} \).
• The imaginary part: derived from \( \frac{\sqrt{- (b^2 - 4ac)}}{2a} \), including the \( i \).

For our quadratic problem, since the discriminant was \(-12\), the roots turned out to be \( 1 + \frac{i\sqrt{3}}{3} \) and \( 1 - \frac{i\sqrt{3}}{3} \). Despite being non-real, these solutions are completely valid and solve the equation within the complex number system, illustrating the power of algebra to handle various types of solutions.