Problem 31 - Intermediate Algebra

Question

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {y=\sqrt{x}} \\ {x^{2}+y^{2}=12} \end{array}\right. $$

Step-by-Step Solution

Answer

The real solution is \((3, \sqrt{3})\).

Step 1: Express y in terms of x

Given the first equation of the system, \( y = \sqrt{x} \). This expresses \( y \) directly in terms of \( x \), which will be used in the second equation.

Step 2: Substitute y into the second equation

Substitute \( y = \sqrt{x} \) into the second equation \( x^2 + y^2 = 12 \). This gives the equation \( x^2 + (\sqrt{x})^2 = 12 \).

Step 3: Simplify the substituted equation

Simplify \( x^2 + (\sqrt{x})^2 = 12 \) to \( x^2 + x = 12 \). This is a quadratic equation in terms of \( x \).

Step 4: Rearrange and solve the quadratic equation

Rearrange the equation \( x^2 + x - 12 = 0 \). Use the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \) with \( a = 1 \), \( b = 1 \), and \( c = -12 \) to solve for \( x \).

Step 5: Apply the Quadratic Formula

Calculate the discriminant: \( b^2 - 4ac = 1^2 - 4(1)(-12) = 49 \). Apply the quadratic formula to find \( x = \frac{{-1 \pm \sqrt{49}}}{2} = \frac{{-1 \pm 7}}{2} \). Therefore, \( x = 3 \) or \( x = -4 \).

Step 6: Determine corresponding y-values

For \( x = 3 \), substitute back into \( y = \sqrt{x} \) to get \( y = \sqrt{3} \). For \( x = -4 \), \( y = \sqrt{-4} \) is not real, hence it is not a valid solution. So, \( (x, y) = (3, \sqrt{3}) \) is the valid real solution.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). The degree of the equation is 2, which means the variable \( x \) is raised to the power of 2. This type of equation is essential because it has a variety of applications in different fields, including physics, engineering, and economics. There are several methods to solve quadratic equations, including:

• Factoring: If the quadratic can be expressed as a product of two binomials, then it can be solved by factoring.
• Completing the Square: Transform the equation into a perfect square trinomial.
• Quadratic Formula: A formula that provides the solution directly by substituting the coefficients \( a \), \( b \), and \( c \) into it.

In our problem, we used the quadratic formula because it is the most direct way to find the solutions for any quadratic equation, especially when factoring is not immediately obvious.

Real Solutions

Real solutions refer to solutions of equations that are real numbers, as opposed to imaginary or complex numbers. In the context of our problem involving nonlinear systems, we are only interested in real solutions because they are applicable to real-world scenarios. For the equation \( x^2 + x = 12 \), the solutions are determined by evaluating the discriminant, \( b^2 - 4ac \), under the square root in the quadratic formula. The discriminant helps determine the nature of the solutions:

• If the discriminant is positive, the equation has two different real solutions.
• If the discriminant is zero, there is one real solution, also known as a repeated root.
• If the discriminant is negative, no real solutions exist, only complex.

In our case, the discriminant was positive (49), resulting in two potential real solutions. However, only one was valid for the given square root condition, \( y = \sqrt{x} \).

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It is given by:\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]This formula works for any quadratic equation and requires substituting the coefficients \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \) into it. The step-by-step use of the quadratic formula involves:

• Calculate the Discriminant: \( b^2 - 4ac \).
• Evaluate the Square Root: Calculate \( \sqrt{b^2 - 4ac} \)
• Plug into the Formula: Use the values in the quadratic formula to solve for \( x \).
• Resolve \( \pm \): Evaluate both the plus and minus options to find the two possible solutions.

In our solution, \( a = 1 \), \( b = 1 \), and \( c = -12 \) were plugged into the formula, leading to valid solutions \( x = 3 \) and \( x = -4 \). Only \( x = 3 \) resulted in a valid real number for \( y \) in the system equation.