Problem 32 - Algebra 2

Question

Solve each equation. $$ x^{4}+6 x^{2}-27=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).

Step 1: Identify a Substitution

Notice that the equation is a type of quadratic in form: it's a quartic (fourth degree) polynomial that can be transformed into a quadratic by substituting a variable for a term. Let us set \( y = x^2 \). Then substitute and rewrite the equation:

Step 2: Rewrite the Equation

Using the substitution \( y = x^2 \), the original equation \( x^4 + 6x^2 - 27 = 0 \) becomes \( y^2 + 6y - 27 = 0 \). Now we have a quadratic in form.

Step 3: Solve the Quadratic Equation

We now solve \( y^2 + 6y - 27 = 0 \) using the quadratic formula: \( y = \frac{{-b \, \pm \, \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = -27 \). Calculate the discriminant: \( b^2 - 4ac = 6^2 - 4(1)(-27) = 36 + 108 = 144 \). Therefore, the roots are \( y = \frac{{-6 \, \pm \, 12}}{2} \).

Step 4: Find the Roots for y

Compute the roots of the quadratic equation. They are \( y = \frac{-6 + 12}{2} = 3 \) and \( y = \frac{-6 - 12}{2} = -9 \). So, \( y = 3 \) and \( y = -9 \).

Step 5: Substitute Back

Remember that \( y = x^2 \). Therefore, \( x^2 = 3 \) and \( x^2 = -9 \). Solve these equations for \( x \).

Step 6: Solve for x

Solve \( x^2 = 3 \), resulting in \( x = \pm \sqrt{3} \). Solve \( x^2 = -9 \). Since we cannot have real solutions for a negative number under the square root in the domain of real numbers, there are no real solutions for \( x^2 = -9 \).

Step 7: Write Final Solutions

The solutions for the original equation are \( x = \sqrt{3} \) and \( x = -\sqrt{3} \), both of which are real numbers.

Key Concepts

Quartic Equation

A quartic equation is a polynomial equation of degree four. They always take the form \[ ax^4 + bx^3 + cx^2 + dx + e = 0 \] where \( a \), \( b \), \( c \), \( d \), and \( e \) are constants, and \( a eq 0 \). Quartic equations can be tricky because of their high degree. They might have up to four real solutions or complex solutions.
One useful technique for dealing with them is recognizing patterns or structures in the equation. For example, some quartic equations can be manipulated into a quadratic form by substitution. This simplification makes them easier to solve using well-known methods such as the quadratic formula. In our example, we restructured \( x^4 + 6x^2 - 27 = 0 \) into \( y^2 + 6y - 27 = 0 \) by substituting \( y = x^2 \). This substitution transforms the challenging quartic problem into a more manageable quadratic equation. This approach allows us to leverage simpler techniques to identify solutions.

Quadratic Formula

The quadratic formula is a powerful tool used to find solutions (roots) to quadratic equations. Once you have an equation in the form \[ ax^2 + bx + c = 0 \], this formula helps find the values of \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
This formula arises from the process of completing the square and allows us to solve any quadratic equation, even when factoring is difficult or impossible. To apply the formula, follow these steps:

• Identify \( a \), \( b \), and \( c \) from your quadratic equation.
• Calculate the discriminant \( b^2 - 4ac \). The discriminant tells you how many real solutions the equation has.
• Plug the values into the quadratic formula itself.
• Simplify and solve for the variable to find the roots.

In the exercise, once the equation \( y^2 + 6y - 27 = 0 \) was recognized as a quadratic equation, the quadratic formula made it possible to find \( y = 3 \) and \( y = -9 \).

Real Solutions

Real solutions refer to numbers that satisfy an equation and are real numbers as well. This means that they are not imaginary numbers (with the imaginary unit \( i \)). Real solutions are the intersections of the equation with the horizontal axis on a graph.
In a polynomial equation, the solutions correspond to the points where the polynomial equals zero or where the graph crosses the x-axis:

• If a polynomial equation has a discriminant greater than zero, it generally has two real solutions.
• If the discriminant equals zero, it has one real solution.
• If the discriminant is less than zero, the solutions are not real; instead, they are complex or imaginary numbers.

In the given problem, the equation \( y^2 + 6y - 27 = 0 \) yields a discriminant of 144, which indicates two real solutions. Thus, we found two \( y \) values: 3 and -9. However, since \( y = x^2 \), only \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) are real for \( x^2 = 3 \), while there are no real solutions for \( x^2 = -9 \).

Polynomial Transformation

Polynomial transformation is a technique where an equation is rewritten or modified to simplify the solution process. This involves manipulating the polynomial into a different form with an easier pathway to finding solutions. It's particularly useful for higher-degree polynomials like quartics.
In the problem \( x^4 + 6x^2 - 27 = 0 \), transforming the polynomial makes a challenging equation more approachable. We used substitution to transform a quartic equation into a quadratic one:

• Identify the repeated pattern or term, as with \( x^2 \).
• Substitute it with a single variable, such as \( y = x^2 \), reducing the degree of the polynomial.
• Solve the modified equation (like we did with the quadratic form \( y^2 + 6y - 27 = 0 \)).

This transformation simplifies the path to finding solutions as we can use established techniques like the quadratic formula to solve what initially seemed a complex quartic equation. This is why polynomial transformations are essential in algebra.