Problem 22 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{l} x^{2}+y^{2}=49 \\ x-2 y^{2}=7 \end{array}$$

Step-by-Step Solution

Answer

The solutions to the given system of equations are Point 1: \((x \approx 15, y \approx 2)\) and Point 2: \((x \approx 15, y \approx -2)\).

Step 1: Solve one equation for one variable

For this given problem, it is easier to solve the second equation for x since it's a simpler expression. Therefore, we'll solve for x in terms of y in the second equation: $$x = 2y^2 + 7$$

Step 2: Substitute the expression for x into the first equation

Now we substitute the expression for x we found in Step 1 into the first equation: \((2y^2 + 7)^2 + y^2 = 49\)

Step 3: Simplify and solve the equation for y

Expand the equation we obtained in Step 2, and then simplify it to obtain a quadratic equation in y: \((4y^4 + 28y^2 + 49) + y^2 = 49\) \(4y^4 + 29y^2 - 49 = 0\) To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula. Since the equation is nontrivial, we will use the quadratic formula: \(y^2 = \frac{-29 + \sqrt{29^2 - 4(4)(-49)}}{8}\) or \(y^2 = \frac{-29 - \sqrt{29^2 - 4(4)(-49)}}{8}\) Solving for y gives: \(y = \sqrt{\frac{-(29 - \sqrt{29^2 - 4(4)(-49)})}{8}}\), \(-\sqrt{\frac{-(29 - \sqrt{29^2 - 4(4)(-49)})}{8}}\), \(\sqrt{\frac{-(29 + \sqrt{29^2 - 4(4)(-49)})}{8}}\), \(\sqrt-\frac{(29 + \sqrt{29^2 - 4(4)(-49)})}{8}\) We need to check the values of y to see which ones are real and valid solutions: After calculating, we get: \(y \approx \pm 2\) Now we have two possible values for y.

Step 4: Calculate the corresponding x values

For each value of y, use the expression we found in Step 1 to calculate the corresponding x values: Case 1: \(y \approx 2\) \(x = 2(2^2) + 7 \Rightarrow x \approx 15\) Case 2: \(y \approx -2\) \(x = 2((-2)^2) + 7 \Rightarrow x \approx 15\) We find that x is the same for both y values.

Step 5: List the solution(s) of the system

After solving the equations, we found two possible solutions for the given system of equations: Point 1: \((x \approx 15, y \approx 2)\) and Point 2: \((x \approx 15, y \approx -2)\) These are the points where the circle and parabola described by the given equations intersect each other.

Key Concepts

Quadratic Equations

Quadratic equations play a crucial role in solving many algebraic problems, such as the system of equations in our exercise. A quadratic equation is generally written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Here, the variable is squared, making the equation non-linear.

Quadratic equations can describe a variety of physical phenomena such as projectile motion or any scenario where the change in a quantity is related to the square of another. The solutions to quadratic equations can be real or complex, depending on the discriminant \(b^2 - 4ac\). If it’s non-negative, we get real solutions; otherwise, they turn out to be complex.

In the example from our exercise, the term \(4y^4 + 29y^2 - 49 = 0\) formed during the solving process represents a type of quadratic in terms of \(y^2\). Understanding the fundamental properties of quadratics helps solve such systems effectively.

• Degree: It's a polynomial of degree 2.
• Roots: Solutions to the equation, which can be found by different methods.
• Parabolic Shape: The graph of a quadratic equation is a parabola, opening upward if \(a > 0\) or downward if \(a < 0\).

Substitution Method

The substitution method is a powerful tool for solving systems of equations, especially when one equation is simpler to manipulate. It involves solving one equation for one variable, and then substituting this expression into the other equation. This method reduces the system into a single equation with one variable.

In our system, we chose to solve \(x - 2y^2 = 7\) for \(x\) since it is a straightforward algebraic manipulation. By rewriting this as \(x = 2y^2 + 7\), we could replace \(x\) in the first equation \(x^2 + y^2 = 49\) and transform it solely in terms of \(y\).

The benefits of the substitution method include:

• It is precise and directs focus on one variable at a time.
• Frequently reduces the complexity of the problem, especially in linear contexts.
• It is straightforward for problems where one variable is already isolated.

However, the substitution method can turn quite algebraically challenging when dealing with higher-degree equations, requiring careful expansion and simplification.

Quadratic Formula

The quadratic formula is a universally applicable method learned in algebra to find solutions, or roots, of a quadratic equation. It is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, the solution is determined for the general form \(ax^2 + bx + c = 0\). This formula arises from completing the square of the quadratic equation.

The term \(b^2 - 4ac\) within the square root is called the discriminant. It indicates the nature and number of solutions:

• Positive discriminant: Two distinct real solutions.
• Zero: Exactly one real solution, or repeated root.
• Negative: Two complex solutions.

In our problem, after converting the system to a quadratic form involving \(y^2\), we applied the quadratic formula to find the values of \(y\). This helped unravel the possible solutions within the context of the original system. Remember to ensure that calculations are precise to determine accurate solutions, especially with roots involving irrational numbers.

Algebraic Solutions

Algebraic solutions involve solving equations and systems analytically to find numerical or symbolic solutions. This involves using techniques like substitution, factoring, or applying the quadratic formula, as illustrated in solving our system.

The primary objective of algebraic solutions is to achieve an answer that satisfies all equations in the system simultaneously. This system involves two equations representing geometric shapes: a circle and a parabola. Their intersection points are where the algebraic solutions give meaningful results.

Key considerations when finding algebraic solutions include:

• Verification: Always substitute back to verify if the proposed solutions satisfy both original equations.
• Real vs Complex: Check which solutions fit within the problem's real-world context.
• Simplification: Simplify expressions to their most manageable form before solving.

For our specific exercise, finding and verifying solutions pinpoint the intersection of the two curves, with solutions such as \((x \approx 15, y \approx 2)\) fitting within the problem constraints. Algebraic methods not only provide solutions but deepen understanding of the relationship between geometric and algebraic representations.