Problem 56 - Algebra and Trigonometry

Question

\(55-58\) m Find all real solutions of the equation, rounded to two decimals. $$ x^{4}-8 x^{2}+2=0 $$

Step-by-Step Solution

Answer

The real solutions are approximately \( x = \pm 2.78 \) and \( x = \pm 0.5 \).

Step 1: Substitute Variable

Let's make a substitution to simplify the problem. Define a new variable, \( y = x^2 \). Then the equation becomes \( y^2 - 8y + 2 = 0 \).

Step 2: Solve the Quadratic Equation

Now we need to solve the quadratic equation \( y^2 - 8y + 2 = 0 \) using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1 \), \( b=-8 \), and \( c=2 \).

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac = (-8)^2 - 4 \cdot 1 \cdot 2 = 64 - 8 = 56 \).

Step 4: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \( y = \frac{-(-8) \pm \sqrt{56}}{2 \times 1} = \frac{8 \pm \sqrt{56}}{2} \).

Step 5: Compute the Roots for y

Simplify \( y = \frac{8 \pm \sqrt{56}}{2} \) to find \( y = \frac{8 + \sqrt{56}}{2} \) and \( y = \frac{8 - \sqrt{56}}{2} \). These yield numerical approximations: \( y_1 \approx 7.75 \) and \( y_2 \approx 0.25 \).

Step 6: Find x Given y Values

Since \( y = x^2 \), solve for \( x \,\) by taking square roots: \( x = \pm\sqrt{y} \). So for \( y_1 = 7.75 \), \( x \approx \pm \sqrt{7.75} \). And for \( y_2 = 0.25 \), \( x \approx \pm\sqrt{0.25} \).

Step 7: Calculate x Values

Calculate \( x \) for each \( y \): \( x_1 \approx \pm 2.78 \) from \( \sqrt{7.75} \), and \( x_2 = \pm 0.5 \) from \( \sqrt{0.25} \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in solving complex equations by simplifying them. When you encounter equations like the one given in the exercise, it is often beneficial to reduce the complexity by introducing a new variable. This is done by substituting a part of the equation with something simpler.

For instance, in the equation \(x^4 - 8x^2 + 2 = 0\), a direct solution could be cumbersome. Instead, we substitute \(y = x^2\), transforming the original equation into \(y^2 - 8y + 2 = 0\), which is a standard quadratic equation.

The idea is straightforward:

• Identify a term that can make the equation simpler.
• Substitute it with a new variable.
• Solve the new and simpler equation.

Once resolved, remember to backtrack and replace the substitution to find the actual solution for the original variable, \(x\), in this case.

Quadratic Formula

The quadratic formula is a reliable method for solving quadratic equations, of the form \(ax^2 + bx + c = 0\). In our exercise, once the substitution is made, we apply the quadratic formula to the equation \(y^2 - 8y + 2 = 0\). This formula provides the solutions for \(y\) by calculating:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = 1\), \(b = -8\), and \(c = 2\). Plugging these into the formula allows us to solve for \(y\). It uses basic operations of addition, subtraction, multiplication, division, and finding a square root, making it universal and efficient.

This formula is essential for every student learning quadratic equations because it guarantees a method to find the solutions when other methods, like factoring, are either too complex or inapplicable.

Discriminant

The discriminant is a crucial component of the quadratic formula, as it helps determine the nature of the roots of a quadratic equation.
In the formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the expression \(b^2 - 4ac\) inside the square root is the discriminant, which we denote as \(D\). In the exercise, for \(y^2 - 8y + 2 = 0\), the discriminant is calculated as:

• \(D = (-8)^2 - 4 \cdot 1 \cdot 2 = 64 - 8 = 56\)

The value of the discriminant tells us about the real and complex nature of the roots:

• If \(D > 0\), there are two distinct real roots.
• If \(D = 0\), there is exactly one real root (a repeated root).
• If \(D < 0\), the equation has complex roots.

In this case, since \(D = 56\), which is greater than zero, it confirms that the equation has two distinct real roots.