Problem 30 - Precalculus: Mathematical 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 solution is \( (x, y) = (-\sqrt{6}, 6) \).

Step 1: Express x in terms of y

Given the first equation: \( x + \sqrt{y} = 0 \), we can isolate \( x \) as follows: \( x = -\sqrt{y} \). Now we have \( x \) expressed in terms of \( y \).

Step 2: Substitute x into the second equation

Take the expression \( x = -\sqrt{y} \) and substitute it into the second equation \( y^2 - 4x^2 = 12 \). Substituting gives: \( y^2 - 4(-\sqrt{y})^2 = 12 \). Simplify this to \( y^2 - 4y = 12 \).

Step 3: Rearrange and solve for y

The equation \( y^2 - 4y = 12 \) can be rearranged to \( y^2 - 4y - 12 = 0 \). This is a quadratic equation in standard form, so solve it using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -4 \), and \( c = -12 \).

Step 4: Apply the quadratic formula

Apply the quadratic formula: \( y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} \). This simplifies to \( y = \frac{4 \pm 8}{2} \).

Step 5: Find the y-values

Calculating the solutions gives \( y = \frac{12}{2} = 6 \) and \( y = \frac{-4}{2} = -2 \). Since \( y = -2 \) gives an invalid result for \( \sqrt{y} \), it is disregarded, so \( y = 6 \).

Step 6: Find the corresponding x-values

Substituting \( y = 6 \) back into the equation \( x = -\sqrt{y} \) gives \( x = -\sqrt{6} \).

Step 7: Validate Solutions

Plug \( x = -\sqrt{6} \) and \( y = 6 \) back into both original equations to ensure they satisfy both.\( x + \sqrt{y} = -\sqrt{6} + \sqrt{6} = 0\) and \( y^2 - 4x^2 = 36 - 4(6) = 12\). Both equations are satisfied.

Key Concepts

Quadratic Equations

Quadratic equations are powerful mathematical tools that model many real-world situations. These equations have the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( x \) represents the variable.Solving quadratic equations typically involves finding the values of \( x \) that make the equation true. There are different methods to solve them, such as:

• Factoring
• Completing the square
• Using the quadratic formula

In our exercise, the quadratic formula is the star. This formula is derived from completing the square and is expressed as:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \( b^2 - 4ac \) is called the discriminant, and it helps to determine the nature of the roots of the quadratic equation. Positive, negative, or zero discriminants indicate the type and number of solutions. This method is particularly useful when the quadratic equation cannot be easily factored.

Substitution Method

The substitution method is an effective way to find solutions to systems of equations. Here, you solve one equation for one variable and then substitute that expression into the other equation.This is especially handy when dealing with equations that contain more complex terms like radicals or squared numbers.In the given system of equations:\[\begin{array}{l}x + \sqrt{y} = 0 \y^2 - 4x^2 = 12\end{array}\]we begin by isolating \( x \) in the first equation:\[ x = -\sqrt{y} \]Then, replace \( x \) in the second equation with \( -\sqrt{y} \), simplifying it to a smaller problem that only involves one variable, \( y \). This method makes complex systems more manageable and often reduces them to a form that is easier to solve, such as a quadratic equation.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. Here's a simple process to remember:

• Identify the type of equation: linear, quadratic, etc.
• Choose an appropriate method: substitution, elimination, factoring, etc.
• Break down the problem into simpler parts if necessary.
• Use algebraic techniques to isolate the variable and solve.

In our system, after substitution, the equation was transformed into a quadratic form:\[ y^2 - 4y = 12 \]Rearranging gives us:\[ y^2 - 4y - 12 = 0 \]Using the quadratic formula, we found the solutions for \( y \). Once \( y \) was determined, we used it to find the value of \( x \) by substitution back into \( x = -\sqrt{y} \).This process highlights how starting with a system of seemingly complex equations can lead to straightforward solutions with the right strategies.