Problem 7 - Algebra 2

Question

Find the exact solutions by using the Quadratic Formula. \(x^{2}+3 x+8=5\)

Step-by-Step Solution

Answer

The solutions are \(x = \frac{-3 + i\sqrt{3}}{2}\) and \(x = \frac{-3 - i\sqrt{3}}{2}\).

Step 1: Move All Terms to One Side

Start by subtracting 5 from both sides of the equation to set it to zero. This will allow us to use the Quadratic Formula.Original equation: \[x^2 + 3x + 8 = 5\]Subtract 5:\[x^2 + 3x + 3 = 0\]

Step 2: Identify Coefficients

In order to use the Quadratic Formula, identify the coefficients of the equation \(ax^2 + bx + c = 0\).Here, \(a = 1\), \(b = 3\), and \(c = 3\).

Step 3: Write the Quadratic Formula

The Quadratic Formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Step 4: Substitute the Coefficients into the Formula

Substitute \(a = 1\), \(b = 3\), and \(c = 3\) into the formula:\[x = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times 3}}{2 \times 1}\]This simplifies to:\[x = \frac{-3 \pm \sqrt{9 - 12}}{2}\]

Step 5: Simplify the Expression Inside the Square Root

Calculate the value under the square root:\[9 - 12 = -3\]Since this is a negative number, we'll have an imaginary number as part of the solution:\[\sqrt{-3} = i\sqrt{3}\] where \(i\) is the imaginary unit.

Step 6: Solve for the Values of x

Substituting back, we get:\[x = \frac{-3 \pm i\sqrt{3}}{2}\]This yields two solutions:\[x = \frac{-3 + i\sqrt{3}}{2}\] and \[x = \frac{-3 - i\sqrt{3}}{2}\]

Key Concepts

Understanding Imaginary Numbers

When working with quadratic equations, you might encounter a situation where you need to take the square root of a negative number. This leads to the concept of imaginary numbers. Imaginary numbers are numbers that can be expressed as the product of a real number and the imaginary unit, denoted by \(i\). The imaginary unit has the crucial property that \(i^2 = -1\).

• For example, the square root of \(-1\) is \(i\).
• Thus, \(\sqrt{-3}\) becomes \(i\sqrt{3}\), as expressed in the solution to the exercise above.

The term 'imaginary' can be a bit misleading. These numbers are very real in terms of their applications in various fields such as engineering and physics. They help in solving equations that don't have a solution in the set of real numbers.

What is a Quadratic Equation?

A quadratic equation is a second-degree polynomial equation in a single variable. It is generally represented in the standard form \(ax^2 + bx + c = 0\) where:

• \(a\), \(b\), and \(c\) are constants, with \(a eq 0\) since if \(a=0\), the equation becomes linear, not quadratic.
• The equation is called 'quadratic' because 'quad' means 'square,' which is the highest power of the variable \(x\).

Quadratic equations can have different types of solutions depending on the value of the discriminant \(b^2 - 4ac\).

• When the discriminant is positive, the equation has two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution.
• However, if the discriminant is negative, as in the previous exercise, the solutions are complex or involve imaginary numbers.

Navigating Complex Solutions

Complex solutions arise when the discriminant of the quadratic equation is negative. This usually results in solutions that involve imaginary numbers.

• A complex number consists of a real part and an imaginary part, which are often expressed in the form \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part.
• In our solved equation, the solutions \(x = \frac{-3 + i\sqrt{3}}{2}\) and \(x = \frac{-3 - i\sqrt{3}}{2}\) represent complex numbers.

Complex solutions occur in conjugate pairs when solving quadratic equations with real coefficients. This means if one root is \(a + bi\), the other is \(a - bi\). Understanding that these solutions appear in pairs helps in verifying whether the solutions obtained are correct and complete.