Problem 32 - Algebra A Combined Function

Question

Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. $$ 3 y^{2}+6 y+5=0 $$

Step-by-Step Solution

Answer

The complex solutions are \(y = -1 \pm \frac{\sqrt{6}i}{3}\).

Step 1: Identify the coefficients

For the given quadratic equation \(3y^2 + 6y + 5 = 0\), identify and note down the coefficients: \(a = 3\), \(b = 6\), and \(c = 5\).

Step 2: Calculate the discriminant

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(\Delta = b^2 - 4ac\). Substitute \(b = 6\), \(a = 3\), and \(c = 5\) to find \(\Delta = 6^2 - 4 \, \times \, 3 \, \times \, 5\). Compute this to get \(\Delta = 36 - 60 = -24\).

Step 3: Analyze the discriminant

Since the discriminant \(\Delta = -24\) is less than zero, this indicates that the solutions for the quadratic equation are complex and not real.

Step 4: Apply the quadratic formula

The quadratic formula is \(y = \frac{-b \pm \sqrt{\Delta}}{2a}\). Substitute \(a = 3\), \(b = 6\), and \(\Delta = -24\) into the formula to get \(y = \frac{-6 \pm \sqrt{-24}}{6}\).

Step 5: Simplify the expression

Since \(\sqrt{-24} = \sqrt{24}i = 2\sqrt{6}i\), substitute back into the quadratic formula to get \(y = \frac{-6 \pm 2\sqrt{6}i}{6}\). Simplify the terms as \(y = -1 \pm \frac{\sqrt{6}i}{3}\).

Key Concepts

Complex Solutions

When solving quadratic equations, sometimes the solutions are not what we call 'real numbers.' Instead, they are complex numbers. A complex number includes a real part and an 'imaginary' part, often expressed with the symbol \(i\). The equation \( 3y^2 + 6y + 5 = 0\) provides an example of this.
In this exercise, after calculating the discriminant, we found it to be negative. This negative value means our solutions will be complex, specifically involving imaginary numbers. Here, we simplified \(\sqrt{-24}\) into \(2\sqrt{6}i\).

• Remember, \(i\) is defined as the square root of \(-1\).
• The presence of \(i\) indicates an imaginary component, turning real numbers into complex solutions.

Look how these complex solutions encompass both the beautiful realm of real numbers and extend beyond into the world of the imaginary.

Discriminant

In quadratic equations, the discriminant plays a crucial role in determining the nature of the solutions. The discriminant is calculated using the formula \(\Delta = b^2 - 4ac\).
For the quadratic equation \(3y^2 + 6y + 5 = 0\), the discriminant was calculated to be \(-24\). This value told us that the solutions were not real, because:

• If \(\Delta > 0\), we get two distinct real solutions.
• If \(\Delta = 0\), we get exactly one real solution, also called a repeated root.
• If \(\Delta < 0\), as is the case here, the solutions are complex and not real.

Understanding the discriminant can help you quickly assess the type of solutions a quadratic equation will have, without needing to fully solve it. It acts like a "Sneak Peek" into the nature of the solutions.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\) where \(a, b,\) and \(c\) are constants, and \(aeq0\). Many real-world problems lead to quadratic equations: from calculating areas to projectile motion. The general technique for solving them involves the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• "\(-b\pm\)" reflects the two possible solutions we usually obtain.
• The term under the square root, \(b^2 - 4ac\), is the discriminant.

Choosing how to approach a quadratic equation often depends on the problem context: sometimes it’s solved by factoring or completing the square, but the quadratic formula provides a reliable solution for any situation.
In our example, after using the quadratic formula, the complex solutions confirm that quadratic equations reach into the rich and diverse fields of mathematical numbers, providing deeper insights and broader applications beyond just numbers on a page.