Problem 71 - Algebra and Trigonometry

Question

\(71-74\) . Solve the equation for the variable \(x\) . The constants \(a\) and \(b\) represent positive real numbers. $$ x^{4}+5 a x^{2}+4 a^{2}=0 $$

Step-by-Step Solution

Answer

No real solution exists for the equation.

Step 1: Substitute a temporary variable

Let us take a temporary variable, \( y \), such that \( y = x^2 \). This simplification transforms the given equation into a quadratic equation in terms of \( y \). Thus, we have: \( y^2 + 5a y + 4a^2 = 0 \).

Step 2: Identify coefficients

This quadratic equation is in standard form with coefficients: \( A = 1 \), \( B = 5a \), and \( C = 4a^2 \).

Step 3: Apply the quadratic formula

Use the quadratic formula \( y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) to find the values of \( y \). Substitute the coefficients into the formula: \[ y = \frac{-5a \pm \sqrt{(5a)^2 - 4 \cdot 1 \cdot 4a^2}}{2 \cdot 1} \].

Step 4: Simplify under the square root

Calculate the discriminant: \( (5a)^2 - 4 \cdot 1 \cdot 4a^2 = 25a^2 - 16a^2 = 9a^2 \).

Step 5: Solve for y

Now, calculate \( y \) using the quadratic formula: \[ y = \frac{-5a \pm \sqrt{9a^2}}{2} = \frac{-5a \pm 3a}{2} \]. This results in \( y_1 = \frac{-5a + 3a}{2} = -a \) and \( y_2 = \frac{-5a - 3a}{2} = -4a \).

Step 6: Back-substitute to find x

Since \( y = x^2 \), it logically follows that \( x^2 = -a \) and \( x^2 = -4a \). However, since \( a \) is positive, both \( -a \) and \( -4a \) are negative, and there are no real solutions for \( x \) because the square of a real number cannot be negative.

Key Concepts

Discriminant

When solving a quadratic equation, determining the discriminant can provide valuable insight. The discriminant is the part of the quadratic formula found under the square root: \( B^2 - 4AC \). In simple terms, it helps us understand the nature of the roots of the quadratic equation.

• If the discriminant is positive, the quadratic has two distinct real roots.
• If it is zero, the quadratic has one real double root.
• If negative, the quadratic has no real roots, indicating complex or imaginary roots.

In the original problem, the discriminant was calculated to be \( 9a^2 \), which is a positive number, suggesting that there are two real and distinct solutions for \( y \). However, when translated back into the context of the original equation involving \( x^2 \), the outcomes are not viable for real values of \( x \) because they are negative, thereby eliminating real solutions.

Quadratic Formula

One of the most powerful tools in algebra for dealing with quadratic equations is the quadratic formula. This formula provides an exact solution to quadratics of the form \( Ax^2 + Bx + C = 0 \) and is written as: \[y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\]This formula takes into account the coefficients of the quadratic equation and applies the discriminant to determine the roots.
Using it involves plugging the calculated coefficients into the formula and then simplifying to find the solutions.
In the exercise provided, \( A = 1 \), \( B = 5a \), and \( C = 4a^2 \) were identified, allowing the calculation of \( y \) using this formula. Despite the success in finding \( y \), when applying the formula's results back to \( x \), no valid real values for \( x \) were possible due to the nature of the solutions being negative.

Substitution Method

The substitution method is a strategic technique used in solving equations that might otherwise be complex. It involves replacing a variable with another to simplify the equation. In the case of non-linear equations, this approach can transform them into a familiar form, such as a quadratic.
In the original exercise, the substitution method was employed by introducing the variable \( y \) as \( x^2 \). This clever substitution reformed the equation from a quartic to a quadratic, making it solvable using the quadratic formula.

• This method helped simplify the problem, making it easier to identify coefficients and apply the quadratic formula effectively.
• The transition from a complex equation to a standard quadratic was crucial for diagnosing the nature of the solutions through the discriminant.

Ultimately, although substituting made the equation manageable, the back-substitution revealed the real nature of \( x \)'s potential values, leading to the conclusion that no real solutions existed due to negative results.