Problem 65 - Algebra and Trigonometry

Question

\(61-70\) . Find all solutions, real and complex, of the equation. $$ x^{4}-6 x^{2}+8=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = 2, -2, \sqrt{2}, -\sqrt{2} \).

Step 1: Recognize the equation's type

Notice that the given equation is a quadratic in form, with $x^4$ and $x^2$ terms. You can make a substitution to simplify it. Let $y = x^2$.

Step 2: Substitute and rewrite

Using the substitution \(y = x^2\), rewrite the equation as a quadratic: \[ y^2 - 6y + 8 = 0 \]

Step 3: Solve the quadratic equation

Solve the quadratic equation \( y^2 - 6y + 8 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=-6, c=8 \). \[ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \] \[ y = \frac{6 \pm \sqrt{36 - 32}}{2} \] \[ y = \frac{6 \pm \sqrt{4}}{2} \] \[ y = \frac{6 \pm 2}{2} \] This gives us the solutions \( y = 4 \) and \( y = 2 \).

Step 4: Reverse substitution

Since \( y = x^2 \), substitute back the values to find \( x \). \( x^2 = 4 \) gives \( x = \pm 2 \). \( x^2 = 2 \) gives \( x = \pm \sqrt{2} \). Thus, the solutions for \( x \) are \( 2, -2, \sqrt{2}, -\sqrt{2} \).

Step 5: Verify possible complex solutions

Since all solutions derived are real numbers and satisfy the original equation, no complex solutions exist in this case. Real solutions suffice.

Key Concepts

Squared Terms Substitution

When working with equations such as \(x^4 - 6x^2 + 8 = 0\), they can often appear intimidating due to the higher degree terms. However, these types of equations may be simplified using the technique known as squared terms substitution. This method can turn what seems like a complex polynomial into a more familiar quadratic form.

Here's how it works:

• Identify that the equation contains terms like \(x^4\) and \(x^2\).
• Recognize that \(x^4\) can be expressed in terms of \(x^2\). Specifically, \(x^4 = (x^2)^2\).
• Make a substitution to simplify the equation. Let \(y = x^2\). This transforms the original equation of \(x^4 - 6x^2 + 8\) into a quadratic equation: \(y^2 - 6y + 8 = 0\).

The primary advantage of squared terms substitution is that it allows you to leverage your skills with quadratic equations to solve more complex polynomial equations. Once the equation is in a quadratic form, you can use familiar methods like factoring, completing the square, or applying the quadratic formula to find solutions.

Real Solutions

After transforming the original equation through substitution, the simplified quadratic equation \(y^2 - 6y + 8 = 0\) can be solved using the quadratic formula. The quadratic formula is given by:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• In this equation, \(a = 1\), \(b = -6\), and \(c = 8\).

Plug these into the formula:

• Calculate the discriminant: \(b^2 - 4ac = (-6)^2 - 4(1)(8) = 36 - 32 = 4\).
• Solve for \(y\): \(y = \frac{6 \pm 2}{2}\), resulting in \(y = 4\) and \(y = 2\).

These solutions for \(y\) translate back to \(x\) with reverse substitution \(x^2 = y\). Therefore, the solutions for \(x\) are:

• \(x^2 = 4\), giving \(x = \pm 2\).
• \(x^2 = 2\), giving \(x = \pm \sqrt{2}\).

All these numbers are real, confirming that the original equation has four real solutions: \(2, -2, \sqrt{2}, -\sqrt{2}\). The existence of a positive discriminant (greater than zero) indicates that all the roots are real.

Complex Solutions

In some cases, a quadratic equation might yield complex solutions. This happens when the discriminant \(b^2 - 4ac\) is less than zero. Complex solutions arise when you are dealing with the square root of a negative number.

• Any square root of a negative number leads to imaginary numbers because the square root of a negative is not defined in the set of real numbers.
• Complex solutions are usually expressed as \(a \pm bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit (\(i = \sqrt{-1}\)).

However, for the equation \(x^4 - 6x^2 + 8 = 0\), the discriminant is \(4\), a positive value. This means all solutions are real, and there are no complex solutions.

Understanding the potential for complex solutions helps with building a deeper understanding of the complete solution set for any quadratic equation. It helps reinforce why certain steps, such as calculating the discriminant, are essential in the quadratic solution process.