Problem 15 - Precalculus: Mathematical for Calculus

Question

Use the elimination method to find all solutions of the system of equations. $$\left\\{\begin{aligned}x-y^{2}+3 &=0 \\\2 x^{2}+y^{2}-4 &=0\end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are (0.5, ±√3.5) and (-1, ±√2).

Step 1: Rearrange the First Equation

Start with the equation \[ x - y^2 + 3 = 0 \]Rearrange it to express \( x \) in terms of \( y \):\[ x = y^2 - 3 \]

Step 2: Substitute into the Second Equation

Substitute \( x = y^2 - 3 \) into the second equation \[ 2x^2 + y^2 - 4 = 0 \]This results in: \[ 2(y^2 - 3)^2 + y^2 - 4 = 0 \].

Step 3: Expand and Simplify

Expand the expression:\[ 2((y^2 - 3)(y^2 - 3)) + y^2 - 4 = 0 \]\[ 2(y^4 - 6y^2 + 9) + y^2 - 4 = 0 \]\[ 2y^4 - 12y^2 + 18 + y^2 - 4 = 0 \]Combine like terms:\[ 2y^4 - 11y^2 + 14 = 0 \].

Step 4: Substitute z = y^2

Let \( z = y^2 \), then the equation becomes:\[ 2z^2 - 11z + 14 = 0 \].

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( z \), where \( a = 2 \), \( b = -11 \), and \( c = 14 \):\[ z = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 14}}{4} \]\[ z = \frac{11 \pm \sqrt{121 - 112}}{4} \]\[ z = \frac{11 \pm 3}{4} \]This gives solutions \( z = 3.5 \) and \( z = 2 \).

Step 6: Determine y Values

Since \( z = y^2 \), solve for \( y \):1. If \( z = 3.5 \), then \( y^2 = 3.5 \) which gives \( y = \pm\sqrt{3.5} \).2. If \( z = 2 \), then \( y^2 = 2 \) which gives \( y = \pm\sqrt{2} \).

Step 7: Calculate Corresponding x Values

Use \( x = y^2 - 3 \) to calculate \( x \) for each \( y \):1. If \( y = +\sqrt{3.5} \), \( x = 0.5 \).2. If \( y = -\sqrt{3.5} \), \( x = 0.5 \).3. If \( y = +\sqrt{2} \), \( x = -1 \).4. If \( y = -\sqrt{2} \), \( x = -1 \).

Step 8: Verify Solutions

Check both equations to confirm that the solutions are correct.1. \((0.5, +\sqrt{3.5})\) and \((0.5, -\sqrt{3.5})\) satisfy both.2. \((-1, +\sqrt{2})\) and \((-1, -\sqrt{2})\) satisfy both.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental mathematical expressions in which the highest power of a variable is squared. A standard form of a quadratic equation looks like this: \( ax^2 + bx + c = 0 \). Here, \( a, b, \) and \( c \) are constants, and \( a eq 0 \). The main objective when solving a quadratic equation is to determine the values of \( x \) that satisfy the equation.
One of the most common methods for solving quadratic equations is using the Quadratic Formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides the solutions by computing possibilities from the expression inside the square root, known as the discriminant. If the discriminant is positive, there are two real solutions. If it is zero, there's exactly one solution, and if it's negative, no real solutions exist (the solutions are complex numbers).
In real-world scenarios, quadratic equations appear regularly. For instance, in calculating projectile motion, determining areas, or optimizing certain operations, you will often find quadratic equations at play.Understanding quadratic equations is crucial because they serve as building blocks for more complex algebraic expressions.

Systems of Equations

Systems of equations consist of multiple equations that share a common set of variables. The main goal is to find values for these variables that satisfy all the equations concurrently. In our example exercise, we are working with two equations that need to be solved together:- \( x - y^2 + 3 = 0 \)- \( 2x^2 + y^2 - 4 = 0 \)Working with systems of equations can be intriguing, as they involve finding common solutions where all conditions given in the equations are satisfied. There are various methods to approach these problems, such as the graphing method, the substitution method, the elimination method, among others.
In this particular example, we use the substitution method to help break down the system. The solution ultimately involves finding the set of \( x \) and \( y \) values that work for both equations, ensuring their correctness by checking each solution in both expressions.

Substitution Method

The substitution method is a practical algebraic tool used to solve systems of equations. By expressing one variable in terms of the others, it simplifies potentially complicated systems into simpler forms that are easier to solve.
Let's consider the system of equations from the exercise. To employ the substitution method, we first manipulate one equation to express one variable explicitly. For instance, rearranging \( x - y^2 + 3 = 0 \) gives us \( x = y^2 - 3 \). Next, we substitute this value into the other equation, transforming the system into a single equation with one variable:- Substitute \( x = y^2 - 3 \) into \( 2x^2 + y^2 - 4 = 0 \), and simplify it.
Doing so leads us to a quadratic equation in terms of \( y \). We further simplify and solve it using classical quadratic techniques (such as factoring or employing the quadratic formula). This reduces our problem to one involving simpler calculations, thereby allowing us to find the values of \( y \) and subsequently \( x \).
By employing this consistent approach, the substitution method ensures accuracy in finding solutions step by step. As you tackle system of equation problems, mastering substitution will prove invaluable in algebra and beyond.