Problem 23 - Intermediate Algebra

Question

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {y=x^{2}+2} \\ {y=-x^{2}+4} \end{array}\right. $$

Step-by-Step Solution

Answer

Solutions are (1, 3) and (-1, 3).

Step 1: Set Equations Equal

Since both equations equal the same variable, \(y\), we can set them equal to each other. This gives us: \[ x^2 + 2 = -x^2 + 4 \]

Step 2: Simplify and Solve for \(x\)

Add \(x^2\) to both sides of the equation to combine like terms: \[ x^2 + x^2 + 2 = 4 \] This simplifies to:\[ 2x^2 + 2 = 4 \] Subtract \(2\) from both sides to isolate the term with \(x\): \[ 2x^2 = 2 \] Divide both sides by \(2\): \[ x^2 = 1 \] Take the square root of both sides to solve for \(x\):\[ x = \pm 1 \]

Step 3: Solve for \(y\) Using \(x=1\)

Substitute \(x = 1\) back into one of the original equations, for instance, \(y = x^2 + 2\):\[ y = 1^2 + 2 = 3 \]

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

Similarly, substitute \(x = -1\) into the same equation:\[ y = (-1)^2 + 2 = 3 \]

Step 5: Identify Solutions

Combine the values of \(x\) and \(y\) obtained in the previous steps to identify the solution pairs:The solutions are \((1, 3)\) and \((-1, 3)\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that involve terms up to the second degree, typically taking the form of \( ax^2 + bx + c = 0 \). These equations can often result in parabolas when graphed on a coordinate plane. The direction and shape of the parabola depend on the coefficients \(a\), \(b\), and \(c\). If \(a > 0\), the parabola opens upwards, resembling a U-shape, whereas if \(a < 0\), it opens downwards. Quadratics are fundamental in algebra as they frequently appear in various problems and scenarios.

Solving quadratic equations can be achieved through various methods such as:

• Factoring, if the equation is easily factorable into the product of binomials.
• Completing the square, which involves rearranging the equation into a perfect square trinomial.
• Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides the roots of the equation directly.

The choice of method often depends on the specific equation and the context in which you encounter it. Quadratic equations can have two real solutions, one real solution, or no real solutions, depending on the value of the discriminant \( b^2 - 4ac \).

Real Solutions

Real solutions refer to values of \(x\) that satisfy an equation and are on the set of real numbers. In the context of quadratic equations, determining whether a real solution exists hinges on the discriminant \(b^2 - 4ac\). This value helps you understand the nature of the solutions:

• If \(b^2 - 4ac > 0\), the quadratic equation has two distinct real solutions.
• If \(b^2 - 4ac = 0\), it has exactly one real solution, known as a repeated or double root.
• If \(b^2 - 4ac < 0\), there are no real solutions; instead, the solutions are complex or imaginary numbers.

This is crucial when solving systems of equations, especially nonlinear systems, as determining real solutions is often the problem's goal. Real solutions provide possible intersections or answers that are viable and interpretable in the given real-world or theoretical context.

Simultaneous Equations

Simultaneous equations are sets of equations with multiple variables that are solved together. The goal is to find common values for variables that satisfy all equations in the system. These can be linear or nonlinear.

In the case of nonlinear systems like the one in this exercise, you often have equations involving powers higher than one or products of variables. The objective is to find where these equations intersect, revealing shared solutions. For example, the equations \( y = x^2 + 2 \) and \( y = -x^2 + 4 \) present parabolic curves, and solving them simultaneously finds their points of intersection.

• The method often involves setting the equations equal to each other, solving for one variable, and substituting back.
• In some cases, graphing can also provide a visual insight into the solutions.

Successfully solving simultaneous equations requires both algebraic manipulation and understanding of the geometric relationships between the functions involved. Through practices like these, students can deeply appreciate the interaction between algebraic expressions and geometric interpretation.