Problem 13 - Precalculus Mathematics 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 \((0.5, \pm \sqrt{3.5})\) and \((-1, \pm \sqrt{2})\).

Step 1: Express x in terms of y from Equation 1

The first equation is \( x - y^2 + 3 = 0 \). Rearrange this equation to express \( x \) in terms of \( y \):\[ x = y^2 - 3 \].

Step 2: Substitute x in Equation 2

Substitute the expression for \( x \) from Step 1 into the second equation: \( 2x^2 + y^2 - 4 = 0 \).\[2(y^2 - 3)^2 + y^2 - 4 = 0.\]

Step 3: Expand and simplify the equation

Expand \((y^2 - 3)^2\):\[ (y^2 - 3)^2 = y^4 - 6y^2 + 9. \]Substitute back:\[2(y^4 - 6y^2 + 9) + y^2 - 4 = 0 \]which simplifies to:\[ 2y^4 - 12y^2 + 18 + y^2 - 4 = 0 \]Combine like terms:\[ 2y^4 - 11y^2 + 14 = 0. \]

Step 4: Substitute back to find x

Let \( y^2 = t \), turning the equation into a quadratic:\[ 2t^2 - 11t + 14 = 0 \].Solve this quadratic equation using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]where \( a = 2 \), \( b = -11 \), \( c = 14 \).

Step 5: Solve quadratic equation

Calculate the discriminant:\[ (-11)^2 - 4\times2\times14 = 121 - 112 = 9 \].The roots are:\[ t = \frac{11 \pm 3}{4} \].The solutions are \( t = 3.5 \) and \( t = 2 \).

Step 6: Determine the solutions for y

Recall \( t = y^2 \). So for \( t = 3.5 \):\[ y^2 = 3.5 \Rightarrow y = \pm \sqrt{3.5} \].For \( t = 2 \):\[ y^2 = 2 \Rightarrow y = \pm \sqrt{2} \].

Step 7: Calculate x for each y value

Substitute back into \( x = y^2 - 3 \).For \( y^2 = 3.5 \), \( x = 3.5 - 3 = 0.5 \).For \( y^2 = 2 \), \( x = 2 - 3 = -1 \).The pairs are:\((x, y) = (0.5, \pm \sqrt{3.5})\) and \((x, y) = (-1, \pm \sqrt{2})\).

Key Concepts

System of Equations

A system of equations is essentially a set of two or more equations with the same variables. The solution to a system of equations is the set of variable values that make all the equations true simultaneously. In many cases, solving these systems can help us describe relationships between different quantities.
Common methods to solve these systems include substitution, elimination, and graphical analysis. In the context of this exercise, the elimination method is specifically highlighted. However, a key step also involves substitution, which will be covered in detail in another section.

• Elimination method: Simplifies the system by eliminating one variable through addition or subtraction.
• Substitution method: Involves expressing one variable in terms of the other using one equation and then substituting this into the other equation.
• Graphical method: Involves graphing each equation on the same set of axes and identifying points of intersection.

For the system of equations in this exercise, the elimination method begins by expressing a variable, but solves it entirely through substitution.

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree two, typically written in the form: \[ ax^2 + bx + c = 0 \]where:

• \(a\), \(b\), and \(c\) are constants and \(a eq 0\)
• The term \(ax^2\) is the quadratic term
• \(bx\) is the linear term
• \(c\) is the constant term

The solutions to a quadratic equation are known as the roots, which can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This exercise led to a quadratic form: \[2t^2 - 11t + 14 = 0\] by substituting \(y^2\) with \(t\). Solving this quadratic equation involved calculating the discriminant, which helps determine the number and type of solutions. Since the discriminant in this case was positive, it indicated two real solutions.

Substitution Method

The substitution method is a strategic way to solve a system of equations. It involves solving one of the equations for a single variable and using the resulting expression to replace the variable in the other equation. This technique can simplify a system of equations into a single equation with one variable.
Here's how it works in the context of the exercise:

• First, we solved the first equation for \(x\): \[ x = y^2 - 3 \]
• Next, this expression for \(x\) was substituted into the second equation: \[ 2(y^2 - 3)^2 + y^2 - 4 = 0 \]

By reducing it to a single-variable equation, we transformed our system into a solvable quadratic equation in terms of \(y\). In turn, this allowed us to delve into the quadratic equation process, a key concept discussed in-depth in another section. This method is not just limited to linear systems but can be applied as seen, to nonlinear systems, including those involving quadratic terms.