Problem 26 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l} x+\sqrt{y}=0 \\ y^{2}-4 x^{2}=12 \end{array}\right.$$

Step-by-Step Solution

Answer

The only real solution is \((-\sqrt{6}, 6)\).

Step 1: Express x in terms of y

From the first equation, we have \( x + \sqrt{y} = 0 \). This can be rewritten as \( x = - \sqrt{y} \). This allows us to express \( x \) in terms of \( y \).

Step 2: Substitute into the second equation

Substitute \( x = - \sqrt{y} \) into the second equation: \( y^{2} - 4(-\sqrt{y})^2 = 12 \). This simplifies to \( y^2 - 4y = 12 \).

Step 3: Simplify the quadratic equation

Rearrange \( y^2 - 4y = 12 \) into standard quadratic form: \( y^2 - 4y - 12 = 0 \).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \) for \( y^2 - 4y - 12 = 0 \), where \( a = 1, b = -4, c = -12 \). Calculate the discriminant: \( (-4)^2 - 4(1)(-12) = 16 + 48 = 64 \). The solutions for \( y \) are \( y = \frac{{4 \pm \sqrt{64}}}{2} = \frac{{4 \pm 8}}{2} \). Thus, \( y = 6 \) or \( y = -2 \).

Step 5: Find corresponding x for each y

Substitute back to find \( x \) for each \( y \). For \( y = 6 \): \( x = -\sqrt{6} \). For \( y = -2 \): \( x = -\sqrt{-2} \), but since \( \sqrt{-2} \) is not real, \( x \) is undefined for real numbers.

Step 6: Verify valid solutions

Check which solutions are valid in the context of real numbers. Since \( y = -2 \) leads to an invalid \( x \) in real numbers, we discard it. Thus, the solution is \( (x, y) = (-\sqrt{6}, 6) \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \(ax^2 + bx + c = 0\). It provides a method for finding the roots or solutions of any quadratic equation, which is an equation that includes a squared term \(x^2\). The formula is given as:

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

This formula is derived from completing the square method and allows you to find the values of \(x\) that satisfy the quadratic equation. The letters \(a\), \(b\), and \(c\) represent the numerical coefficients of the terms.
The term under the square root \(b^2 - 4ac\) is called the discriminant. It indicates the nature of the roots of the equation. A positive discriminant means there are two distinct real roots, a discriminant of zero means there's exactly one real root, and a negative discriminant means no real roots are available; only complex or imaginary roots exist.
Understanding the quadratic formula is crucial when dealing with second-degree polynomials, as it provides a quick and reliable way to ascertain all possible solutions.

Real Number Solutions

When solving equations, finding real number solutions is often the goal, especially in problems involving systems of equations like the one given. Real numbers include all values that can appear on the continuous number line, comprising both rational and irrational numbers.

• Rational numbers can be expressed as fractions, such as \(\frac{1}{2}\) or 5.
• Irrational numbers, such as \(\sqrt{2}\) or \(\pi\), cannot be expressed as simple fractions.

For an equation to have real solutions, any operation such as taking square roots or other similar functions must result in a real number. For example, the equation \(y = -2\) led to an attempt to compute \(x = -\sqrt{-2}\), which isn't possible with real numbers due to the square root of a negative number being undefined in the real number system.
Hence, the solution \((x, y) = (-\sqrt{6}, 6)\) is a valid real solution for the initial problem as it consists of real numbers for both \(x\) and \(y\). This solution was obtained by discarding the non-real possibilities encountered during the solving process.

Square Roots in Equations

Square roots frequently appear in algebra, especially when solving systems of equations or dealing with quadratic expressions. The square root of a number \(x\), expressed as \(\sqrt{x}\), is a value that, when multiplied by itself, gives \(x\). It is essential to keep in mind:

• Square roots of non-negative numbers are real and yield two possible values: a positive and a negative root, due to the property \((\sqrt{x})^2 = x\).
• Square roots of negative numbers do not yield real results within the real number system but instead introduce imaginary numbers.

In our problem, by solving \(x = -\sqrt{y}\), we were able to express \(x\) in terms of \(y\). This helped simplify the problem and resolve it further.
When calculating square roots as part of solving equations, always verify that they lead to meaningful solutions, especially if constraints like only real number solutions apply. Real solutions necessitate that any roots derived must apply to actual values within the established real constraints, ensuring the solution maintains consistency with real-world applications.