Problem 69 - Algebra for College Students

Question

Find all real and imaginary solutions to each equation. $$a^{4}+6 a^{2}+8=0$$

Step-by-Step Solution

Answer

The solutions are \(\pm i\sqrt{2}\) and \(\pm 2i\).

Step 1: Substitute and Simplify

Let us substitute to make the equation easier to solve. Set \[b = a^2\] This gives us the equation:\[b^2 + 6b + 8 = 0\]

Step 2: Solve the Quadratic Equation

Now factor the quadratic equation \[b^2 + 6b + 8 = 0\] We get:\[(b + 2)(b + 4) = 0\] So, the solutions for \(b\) are \[b = -2\] and \[b = -4\]

Step 3: Substitute Back to Find \(a\)

Recall that \(b = a^2\). Thus, we need to solve \[a^2 = -2\] and \[a^2 = -4\].Taking the square root of both equations (and remembering the imaginary unit \(i\)) gives us:For \[a^2 = -2\]\[a = \pm \sqrt{-2} = \pm i\sqrt{2}\]For \[a^2 = -4\]\[a = \pm \sqrt{-4} = \pm 2i\]

Key Concepts

Imaginary Solutions

When solving quadratic equations, we often encounter cases where the solutions are not real numbers. These are known as imaginary solutions. They occur when we take the square root of a negative number.
The imaginary unit, represented as \(i\), is defined as \(i = \sqrt{-1}\). Consequently, the square root of any negative number can be expressed using \(i\). For example:
\[ \sqrt{-2} = i\sqrt{2} \]
\[ \sqrt{-4} = 2i \] In these cases, the solutions involve imaginary numbers because real numbers squared are never negative. Therefore, when the quadratic equation includes squares of negative numbers, the solutions must be imaginary.

Real Solutions

Quadratic equations can also have real solutions. These occur when the quadratic expression can be factored into two real number solutions or when the discriminant (the part of the quadratic formula under the square root) is non-negative.
For the quadratic equation \[ ax^2 + bx + c = 0 \] the discriminant is given by \[ b^2 - 4ac \].
If \[ b^2 - 4ac \] is zero or positive, the equation has real solutions.
Let's look at an example. For the equation \[ a^2 + 6a + 8 = 0 \], we can factor it as \[ (a + 2)(a + 4) = 0 \]. The solutions are real and are \[ a = -2 \] and \[ a = -4 \]. These are straightforward real numbers.

Factoring

Factoring is a method of breaking down a polynomial into simpler 'factors' that, when multiplied together, give back the original polynomial. It's a key method for solving quadratic equations.
Consider the quadratic expression \[ b^2 + 6b + 8 \]. To factor this, we look for two numbers that multiply to \ b^2 \ (the leading coefficient) and add up to \ 6b \ (the middle term coefficient).
Here, the factors are \ (b + 2) \ and \ (b + 4) \ because:
\[ (b + 2)(b + 4) = b^2 + 4b + 2b + 8 = b^2 + 6b + 8 \] Once factored, we set each factor equal to zero: \[ b+2=0 \] and \[ b+4=0 \].
Solving these gives us \ b = -2 \ and \ b = -4 \ as the solutions.
By factoring, complex polynomial equations are simplified, making them easier to solve.