Problem 43 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} y=x^{2}-4 \\ x^{2}-y^{2}=-16 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((0, -4)\), \((3, 5)\), and \((-3, 5)\).

Step 1: Substitute for y

From the first equation, we have \( y = x^2 - 4 \). Use this expression for \( y \) in the second equation.

Step 2: Substitute y into the second equation

Replace \( y \) in the second equation, \( x^2 - y^2 = -16 \), with \( x^2 - 4 \): \[x^2 - (x^2 - 4)^2 = -16.\]

Step 3: Expand and simplify

Begin by expanding the expression \((x^2 - 4)^2\): \[(x^2 - 4)^2 = (x^2 - 4)(x^2 - 4) = x^4 - 8x^2 + 16.\]Replace back to get: \[x^2 - (x^4 - 8x^2 + 16) = -16.\]

Step 4: Simplify the equation

Subtract the expanded quadratic expression from \( x^2 \): \[ x^2 - x^4 + 8x^2 - 16 = -16.\]Simplifying further: \[ -x^4 + 9x^2 = 0.\]

Step 5: Factor the simplified equation

Take out the common factor \( x^2 \): \[-x^2(x^2 - 9) = 0.\]This implies two possible factors: \(-x^2 = 0\) or \(x^2 - 9 = 0\).

Step 6: Solve each factor

Solve each factor separately:1. \(-x^2 = 0\) simplifies to \(x = 0\).2. \(x^2 - 9 = 0\) can be factored or solved as a quadratic giving \(x = 3\) or \(x = -3\).

Step 7: Find corresponding values of y

Substitute each value of \(x\) back to the equation \(y = x^2 - 4\) to find corresponding \(y\).1. For \(x = 0\), \(y = 0^2 - 4 = -4\).2. For \(x = 3\), \(y = 3^2 - 4 = 9 - 4 = 5\).3. For \(x = -3\), \(y = (-3)^2 - 4 = 9 - 4 = 5\).

Step 8: Verify solutions

Verify each \((x, y)\) pair by substituting back into the original equation \(x^2 - y^2 = -16\):1. For \((0, -4)\): \(0^2 - (-4)^2 = -16\) ⇾ \(0 - 16 = -16\), true.2. For \((3, 5)\): \(3^2 - 5^2 = -16\) ⇾ \(9 - 25 = -16\), true.3. For \((-3, 5)\): \((-3)^2 - 5^2 = -16\) ⇾ \(9 - 25 = -16\), true.All solutions satisfy the original system.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree 2. They are usually in the standard form: \( ax^2 + bx + c = 0 \). Quadratic equations have at most two solutions, as their graphs are parabolas that can intersect the x-axis at two points, one point, or not at all. However, even when a parabola doesn't intersect the x-axis, solutions may exist in the set of complex numbers.

• The coefficient \( a \) represents the term that controls the parabola's shape.
• The coefficient \( b \) affects the position of the symmetry axis.
• The constant \( c \) is the y-intercept of the parabola.

Quadratic equations are pivotal in many areas of mathematics due to their applications in physics and engineering, such as calculating projectile paths or electrical currents. Understanding their structure and solutions is therefore essential. You can solve quadratic equations using various methods, the most common being factoring, completing the square, and the quadratic formula. In this exercise, understanding the definition of the quadratic terms helps set up for successful substitution and factoring.

Substitution Method

The substitution method is a technique for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This method simplifies the system to a single equation in one variable, making it easier to solve. In the context of this exercise, solve the first equation for \( y \) in terms of \( x \), knowing \( y = x^2 - 4 \). Substitute this expression into the second equation. This step reduces the system containing two equations into a single polynomial equation. The substitution method is particularly useful when one equation is already solved for a variable, or can be easily solved, facilitating straightforward replacement into other equations. This strategy resolves the complexity of having multiple variables to manage simultaneously.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into simpler "factor" polynomials that, when multiplied together, return the original polynomial. This approach is essential in simplifying expressions and solving equations, particularly quadratic equations. In this exercise, once the single equation for \( x \) is simplified, factoring helps to solve for the variable \( x \).

• Identify common factors first. In this problem: \( -x^4 + 9x^2 = 0 \) can be factored by taking \( x^2 \) common, leading to \( -x^2(x^2 - 9) = 0 \).
• The next step is to factor the remaining simple terms. Here, \( x^2 - 9 \) is recognized as a difference of squares and further factored as \( (x - 3)(x + 3) \).

Factoring is a formative step in solving polynomial equations because it reduces the equation into its simplest components, making it straightforward to find the roots. Understanding and identifying different types of factoring, such as grouping and using formulas like the difference of squares, is crucial for solving polynomial equations efficiently. By breaking down complex expressions through factoring, solving equations and understanding polynomial structures becomes much more accessible.