Problem 25 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{aligned} x^{2} y &=16 \\ x^{2}+4 y+16 &=0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

There are no real solutions to the system of equations.

Step 1: Solve the second equation for y

The second equation is \(x^2 + 4y + 16 = 0\). We can isolate \(y\) by subtracting \(x^2\) and 16 from both sides, and then dividing by 4: \[ 4y = -x^2 - 16 \] \[ y = -\frac{x^2+16}{4} \]

Step 2: Substitute the expression for y into the first equation

Now that we have \(y = -\frac{x^2 + 16}{4}\), substitute this expression into the first equation \(x^2 y = 16\): \[ x^2 \left(-\frac{x^2 + 16}{4}\right) = 16 \] which simplifies to \[-\frac{x^4 + 16x^2}{4} = 16 \].

Step 3: Solve the resulting equation for x

Simplify \(-\frac{x^4 + 16x^2}{4} = 16\) by multiplying both sides by 4 to eliminate the fraction: \[-x^4 - 16x^2 = 64\]. Rearrange to get \(x^4 + 16x^2 + 64 = 0\).

Step 4: Factor the quadratic form

Notice this resembles a quadratic equation in form \(x^4 + 16x^2 + 64 = 0\). Let \(z = x^2\), which gives \(z^2 + 16z + 64 = 0\). This is a perfect square: \((z + 8)^2 = 0\). Therefore, \(z = -8\).

Step 5: Analyze the solution for x

Since \(z = x^2\) and \(z = -8\), we note there are no real solutions for \(x\) since a square must be non-negative. Thus, \(x^2 = -8\) gives no real roots.

Key Concepts

Quadratic Form

Quadratic forms are expressions where variables are raised to a power of two. These equations often appear in the form of

• \(ax^2 + bx + c = 0\),

with different coefficients.
In our problem, the term \(x^4 + 16x^2 + 64 = 0\) is specifically a polynomial that resembles a quadratic due to its structured format.
To solve these, a clever technique is employed where we substitute \(z = x^2\) to simplify the equation into a more recognizable quadratic form:

• \(z^2 + 16z + 64 = 0\).

This method allows us to handle higher-order polynomials easily.

Substitution Method

The substitution method is a powerful tool in solving systems of equations.
Its main idea is to solve one equation for a single variable and then substitute this expression into another equation.
In the given exercise, after isolating \(y\) from the second equation, we substitute it into the first equation.
This step-by-step substitution simplifies the complex system into a single variable equation.
This approach reduces the complexity by transforming multiple equations into a single polynomial form, making it easier to solve.

No Real Solutions

A crucial aspect of understanding equations is recognizing when no real solutions exist.
As seen in the exercise, solving

• \(z = x^2 = -8\)

provides a key insight.
Since the square of a real number cannot be negative, it concludes that these equations have no intersection in the real number domain.
This awareness helps avoid needless calculations and leads to the correct conclusion that the system has no real solutions.

Factoring Equations

Factoring is a vital method to solve quadratic equations, relying on expressing the equation as a product of simpler polynomials.
In this exercise, recognizing

• \((z + 8)^2 = 0\)

denotes a perfect square trinomial, making \(z\) equal to \(-8\).
Factoring is effective because it reduces polynomials into roots that can be easily solved.
This decomposition of the quadratic into identifiable pieces simplifies finding the roots, though, as highlighted, reveals there are no real solutions.
In many cases, this reveals significant properties of the equation such as the lack of real solutions, as seen in this particular case.