Problem 26 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The solutions are \((x, y) = (\sqrt{6}, 6)\) and \((-\sqrt{6}, 6)\).

Step 1: Introduce Variables and Equations

The system of equations given is \(x + \sqrt{y} = 0\) and \(y^2 - 4x^2 = 12\). Our task is to find values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Solve for y in the First Equation

The first equation is \(x + \sqrt{y} = 0\). Rearrange to express \(\sqrt{y}\) in terms of \(x\): \(\sqrt{y} = -x\). Now, square both sides to solve for \(y\): \(y = x^2\).

Step 3: Substitute Expression into Second Equation

Substitute \(y = x^2\) into the second equation \(y^2 - 4x^2 = 12\): \\((x^2)^2 - 4x^2 = 12\), which simplifies to \(x^4 - 4x^2 = 12\).

Step 4: Substitution Simplification

Let \(z = x^2\), transforming the equation into a quadratic: \(z^2 - 4z = 12\). Rearrange to standard quadratic form: \(z^2 - 4z - 12 = 0\).

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \(z^2 - 4z - 12 = 0\) using the quadratic formula: \\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-4\), \(c=-12\). \Calculate the discriminant: \(b^2 - 4ac = (-4)^2 - 4 \times 1 \times (-12) = 64\). \Find the roots: \(z = \frac{4 \pm 8}{2}\), yielding \(z = 6\) and \(z = -2\).

Step 6: Determine Possible x Values

Since \(z = x^2\), we consider \(x^2 = 6\) and ignore \(x^2 = -2\) because the square of a real number cannot be negative. \\(x^2 = 6\) leads to \(x = \pm \sqrt{6}\).

Step 7: Find Corresponding y Values

For \(x = \sqrt{6}\), substitute into \(y = x^2\) to get \(y = 6\). Similarly, for \(x = -\sqrt{6}\), substituting into \(y = x^2\) also gives \(y = 6\).

Step 8: Verify Solutions

Check both pairs \((x, y) = (\sqrt{6}, 6)\) and \((-\sqrt{6}, 6)\) in original equations: \- For \((\sqrt{6}, 6)\), both \(x + \sqrt{y} = 0\) and \(y^2 - 4x^2 = 12\) are satisfied. \- Similarly, for \((-\sqrt{6}, 6)\), both equations are satisfied. Thus, both are solutions.

Key Concepts

Quadratic Equation

In mathematics, a quadratic equation is any equation that can be expressed in the standard form: \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The solutions to a quadratic equation are the values of \(x\) that satisfy the equation. These values can be found by factoring, using the quadratic formula, or by completing the square.

• Factoring involves expressing the quadratic expression as a product of two binomials.
• The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), provides a way to directly find the roots by inserting the values of \(a\), \(b\), and \(c\).
• Completing the square involves rearranging the equation so that one side is a perfect square trinomial.

Quadratic equations often appear in problems involving areas, projectile trajectories, and various optimization problems.
In our exercise, we transformed an expression into a quadratic form: \(z^2 - 4z - 12 = 0\), where \(z = x^2\). We then used the quadratic formula to find \(z\), ultimately leading to the solution of the system of equations.

Real Numbers

Real numbers are all the numbers that can be found on the number line. This includes both rational numbers, such as integers and fractions, and irrational numbers, numbers that cannot be written as a simple fraction.

• Real numbers include zero, positive and negative whole numbers, and numbers with decimals or fractions.
• The set of real numbers is denoted by the symbol \(\mathbb{R}\).

In the context of solving equations, real numbers provide a complete set to work with when determining solutions.
In our quadratic equation exercise, we consider only the real number solutions. Specifically, we evaluated \(x^2 = 6\), leading to real solutions \(x = \pm \sqrt{6}\). Importantly, when encountering \(x^2 = -2\), we disregarded it as no real number squared can give a negative result.

Solution Verification

Solution verification is an essential step in mathematics to ensure that the values found actually satisfy the original equations. It involves substituting the solutions back into the original equations to check their validity.

• Verification helps identify any calculation errors or incorrect assumptions during solving.
• Each solution is substituted into the original system of equations to ensure all conditions are met.

In our exercise, after finding the potential solutions \((x, y) = (\sqrt{6}, 6)\) and \((-\sqrt{6}, 6)\), we verified them by substituting back into both equations: \(x + \sqrt{y} = 0\) and \(y^2 - 4x^2 = 12\).
For both pairs of \((x, y)\), the equations were satisfied, confirming the correctness and completeness of our solutions.