Problem 24 - Intermediate Algebra

Question

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {x=-y^{2}-3} \\ {x=y^{2}-5} \end{array}\right. $$0

Step-by-Step Solution

Answer

The real solutions are \((-4, 1)\) and \((-4, -1)\).

Step 1: Set the Equations Equal to Each Other

Since both equations are equal to \(x\), you can set them equal to each other.\[-y^2 - 3 = y^2 - 5\]

Step 2: Simplify the Equation

Combine like terms to simplify the equation. Add \(y^2\) to both sides:\[-3 = 2y^2 - 5\]Then, add \(5\) to both sides:\[2 = 2y^2\]

Step 3: Solve for \(y\)

Divide both sides by \(2\) to solve for \(y^2\):\[y^2 = 1\]Take the square root of both sides to find \(y\):\[y = \pm 1\]

Step 4: Solve for \(x\) Using \(y = 1\)

Substitute \(y = 1\) into either equation, for example, the first one:\[x = -(1)^2 - 3 = -4\]So, the solution for \(y = 1\) is \((x, y) = (-4, 1)\).

Step 5: Solve for \(x\) Using \(y = -1\)

Substitute \(y = -1\) into either equation, for example, the first one:\[x = -(-1)^2 - 3 = -4\]So the solution for \(y = -1\) is \((x, y) = (-4, -1)\).

Step 6: Verify Real Solutions

Both solutions \((-4, 1)\) and \((-4, -1)\) satisfy both original equations, thereby confirming they are the correct real solutions.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest exponent of the variable is 2. These equations generally take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not zero. They can describe a parabola when graphically represented. Right in this node, a critical thing to remember is:

• Quadratic equations can have one, two, or no real solutions.
• These solutions can be obtained by methods like factoring, completing the square, or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

In our exercise, the equations \(x = -y^2 - 3\) and \(x = y^2 - 5\) are simplified forms of quadratic equations where \(y^2\) acts as the main quadratic element. The equations show a relationship where \(x\) depends on the squared value of another variable, \(y\). Once you solve for \(y^2\), you can easily find corresponding \(x\) values that satisfy the equation in its original form. In simple terms, the variable \(x\) is derived from these quadratic expressions the same way the solutions of traditional quadratic equations are derived.

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In solving a system of equations, the goal is to find the values of the variables that satisfy all equations simultaneously. The systems can be linear or nonlinear, depending on the equations themselves.
The given problem consists of nonlinear equations because the variables involve quadratic terms \(y^2\). Solving a nonlinear system can be a bit more intricate compared to linear systems due to the curvatures and additional solutions that quadratic terms introduce:

• To find a solution, you often use substitution or elimination methods tailored to deal with nonlinear terms.
• The intersection of equations in such a system can result in multiple, single, or no real solutions depending on the nature of the equations and the variables involved.

In this particular exercise, the system \(\{x = -y^2 - 3, x = y^2 - 5\}\) was set equal to each other for \(x\) to eliminate it and reduce to a single quadratic equation, which was solved to find \(y\). This step facilitated reaching values that solve both initial equations efficiently by substituting \(y\) values back.

Real Solutions

Real solutions refer to the set of solutions that are real numbers, as opposed to complex or imaginary numbers. In mathematics, particularly when dealing with quadratic equations, the number of real solutions is determined by the discriminant \(b^2 - 4ac\). If the discriminant is greater than or equal to zero, real solutions exist.
In the context of systems of equations:

• The solutions to the system are considered real if you can substitute these back into the original equations and satisfy them as true statements.
• These solutions often come in pairs if the variable was squared, as in the case of \(y = \pm 1\) from our exercise.

In our nonlinear system, both solutions \((-4, 1)\) and \((-4, -1)\) are categorized as real solutions because they satisfy both original equations. Taking into account the relationship defined, there didn't seem to be any complex numbers involved in our solutions, confirming them to be pure real. This check is fundamental because it validates that all derived solutions indeed align with the conditions laid out by the original polynomial system.