Problem 53 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation. $$4 x^{4}+25 x^{2}+36=0$$

Step-by-Step Solution

Answer

There are no real solutions for the equation.

Step 1: Identify the Quadratic Structure

Notice that the equation \(4x^4 + 25x^2 + 36 = 0\) can be thought of as a quadratic in terms of \(x^2\). By letting \(y = x^2\), the equation becomes \(4y^2 + 25y + 36 = 0\).

Step 2: Apply the Quadratic Formula

For the quadratic equation \(ay^2 + by + c = 0\), the solutions can be found using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \(a = 4\), \(b = 25\), and \(c = 36\). Substitute these values into the formula.

Step 3: Calculate the Discriminant

The discriminant \(b^2 - 4ac\) helps determine the nature of the roots. Calculate it as follows: \(b^2 - 4ac = 25^2 - 4 \times 4 \times 36 = 625 - 576 = 49\). Since the discriminant is positive, there will be two real roots.

Step 4: Find the Roots for \(y\)

Substitute the discriminant and coefficients back into the quadratic formula: \[ y = \frac{-25 \pm 7}{8} \]. This provides two solutions: \(y = \frac{-25 + 7}{8} = -2.25\) and \(y = \frac{-25 - 7}{8} = -4\).

Step 5: Solve for \(x\)

Recall that \(y = x^2\). Suppose \(x^2 = -2.25\) or \(x^2 = -4\). Since both are negative, there are no real solutions for \(x\) because the square roots of negative numbers are not real.

Key Concepts

Quadratic Formula

When faced with a quadratic equation of the form \(ax^2 + bx + c = 0\), a powerful tool to find the solutions is the Quadratic Formula. This formula is expressed as:
\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
It allows you to solve for \(y\) by simply plugging in the coefficients \(a\), \(b\), and \(c\) from the quadratic equation. Here is a quick breakdown of how to use it:

• Identify the coefficients \(a\), \(b\), and \(c\) from your equation.
• Calculate the value of the discriminant \(b^2 - 4ac\).
• Substitute these values into the Quadratic Formula.

This straightforward application enables you to find the roots or solutions of any quadratic equation. Just remember, the equation needs to be quadratic, and all terms should be properly aligned with the formula format.

Discriminant

The discriminant, represented as \(b^2 - 4ac\), plays a crucial role in determining the nature of the roots for quadratic equations. Here's how it works:

• If the discriminant is positive, like in the example \(b^2 - 4ac = 49\), the quadratic equation will have two distinct real roots.
• If the discriminant equals zero, there is exactly one real root, which is also known as a repeated or double root.
• If the discriminant is negative, there are no real roots, but rather two complex roots.

In our example, the discriminant was 49, indicating two real roots for \(y\). However, both potential solutions for \(x^2 = y\) are negative, leading to no real solutions for \(x\). This highlights the importance of understanding the discriminant thoroughly as you navigate through quadratic equations.

Polynomial Equations

Polynomial equations involve terms that are non-negative integer powers of the variable, like \(x^4\) or \(x^2\). In the given exercise, the equation is initially in a quartic form:
\[4x^4 + 25x^2 + 36 = 0\]
To solve this kind of polynomial, identifying a pattern that allows it to resemble a more familiar form, like a quadratic equation, is essential. By substituting \(y = x^2\), the polynomial is transformed into a quadratic:
\[4y^2 + 25y + 36 = 0\]
This transformation is a powerful technique to simplify complex polynomial equations. With the resulting quadratic, standard techniques like the Quadratic Formula can be applied. Always remember:

• Transform complex polynomials into known formats.
• Use substitution effectively to make solving easier.
• Reinterpret the solutions back to the original variables if substitutions were made.

These insights help break down polynomial equations into manageable parts, making solutions more approachable.