Problem 30 - Algebra and Trigonometry

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 solution is \((x, y) = (-\sqrt{6}, 6)\).

Step 1: Substitute for x

From the first equation, solve for \(x\): \(x + \sqrt{y} = 0\). This implies \(x = -\sqrt{y}\). Substitute \(x = -\sqrt{y}\) into the second equation.

Step 2: Substitute in the Second Equation

Replace \(x\) in the second equation \(y^2 - 4x^2 = 12\) with \(-\sqrt{y}\). This gives \(y^2 - 4(-\sqrt{y})^2 = 12\). Simplify the equation to yield \(y^2 - 4y = 12\).

Step 3: Rearrange and Factor

Take \(y^2 - 4y = 12\) and rearrange it into \(y^2 - 4y - 12 = 0\). Now factor the quadratic equation to find \(y\).

Step 4: Solve the Quadratic Equation

The equation \(y^2 - 4y - 12 = 0\) can be solved using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -12\). Calculate the discriminant: \(b^2 - 4ac = 16 + 48 = 64\). Thus \(y = \frac{4 \pm \sqrt{64}}{2}\).

Step 5: Calculate y Values

Solve \(y = \frac{4 \pm 8}{2}\), which gives \(y = 6\) or \(y = -2\). Since \( \sqrt{y}\) is not defined for negative values in the real number system, \(y = -2\) is not acceptable. Therefore, \(y = 6\).

Step 6: Find Corresponding x

Substitute \(y = 6\) back into \(x = -\sqrt{y}\) to find \(x\). This yields \(x = -\sqrt{6}\).

Key Concepts

Quadratic Equation

A quadratic equation is a central topic in algebra that you'll frequently encounter in solving systems of equations. A typical quadratic equation is written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This kind of equation forms a parabola when graphed on the coordinate plane.

• The standard way to solve a quadratic equation is by using factoring, completing the square, or the quadratic formula.
• The quadratic formula is often the most versatile as it applies regardless of whether the quadratic can be easily factored: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\]This formula comes in handy when other methods are cumbersome or impossible.
• In the provided exercise, the equation \(y^2 - 4y - 12 = 0\) is a quadratic equation.

Substitution Method

The substitution method is a valuable technique to solve systems of equations. It's often used when one equation can be easily solved for one variable, which is then substituted back into another equation.

• In the given exercise, we initially solve for \(x\) in terms of \(y\) using the first equation: \(x + \sqrt{y} = 0\), which simplifies to \(x = -\sqrt{y}\).
• We then employ this expression for \(x\) in the second equation to get a single equation in terms of just \(y\).
• This substitution simplifies the problem, allowing us to turn a system of equations into a singular, more manageable equation.

This technique can simplify complex problems and is fundamental in solving many systems of equations, providing a clear path to the solution.

Solution of Equations

Solving equations is about finding the values of the variables that satisfy the given equations. In the case of systems of equations, you need to find values that satisfy all the equations simultaneously.

• For our particular problem, the solutions consist of finding \(x\) and \(y\) that make both equations true.
• By utilizing techniques such as substitution and the quadratic formula, we extract possible solutions for \(y\), which is critical because not all will be valid under all circumstances.
• After finding \(y\), substitute it back to find the respective \(x\) value. In some cases, further verification may be necessary, especially in more complex systems.

Remember, solutions must be checked for validity against all conditions specified by the original problem to ensure they are correct.

Real Number System

The real number system is a fundamental concept in mathematics and encompasses all numbers that can appear on a number line. This includes both rational numbers (like fractions) and irrational numbers (like roots and π).

• When dealing with square roots as in our exercise, the expression \(\sqrt{y}\) imposes restrictions on \(y\) such that only non-negative values make sense within the real numbers. This is because the square root of a negative number isn't defined in the real number system.
• Thus, while solving the quadratic equation, we discard any potential negative solutions for \(y\) to avoid non-real outcomes.
• The need to restrict solutions to real numbers is common in many problems, guiding us to discard certain theoretical solutions that don't fit in the given numerical system.

Adhering to these constraints ensures that the solutions are valid within the context of the problem and adheres to the real-world models we are often trying to solve.