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 \((0.5, \sqrt{3.5} ), (0.5, -\sqrt{3.5} ), (-1, \sqrt{2}), (-1, -\sqrt{2}) \).

Step 1: Solve the first equation for x

The first equation is \( x - y^2 + 3 = 0 \). To solve for \( x \), rearrange the terms to get \( x = y^2 - 3 \).

Step 2: Substitute x in the second equation

Substitute \( x = y^2 - 3 \) into the second equation \( 2x^2 + y^2 - 4 = 0 \). This gives us \( 2(y^2 - 3)^2 + y^2 - 4 = 0 \).

Step 3: Expand and simplify the equation

Expand \( (y^2 - 3)^2 \) to get \( y^4 - 6y^2 + 9 \). The equation becomes \( 2(y^4 - 6y^2 + 9) + y^2 - 4 = 0 \). Simplify to get \( 2y^4 - 12y^2 + 18 + y^2 - 4 = 0 \).

Step 4: Combine like terms

Combine like terms to simplify the equation to \( 2y^4 - 11y^2 + 14 = 0 \).

Step 5: Solve the quadratic in terms of y²

Let \( z = y^2 \). Then the equation becomes \( 2z^2 - 11z + 14 = 0 \). Solve this quadratic equation using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2 \), \( b = -11 \), \( c = 14 \).

Step 6: Calculate the roots of the quadratic

The discriminant \( b^2 - 4ac \) is \( (-11)^2 - 4 \cdot 2 \cdot 14 = 121 - 112 = 9 \). So, the roots are \( z = \frac{11 \pm 3}{4} \). Calculate to find \( z = 3.5 \) or \( z = 2 \).

Step 7: Find values of y

Since \( z = y^2 \), if \( z = 3.5 \) then \( y = \pm \sqrt{3.5} \), and if \( z = 2 \) then \( y = \pm \sqrt{2} \).

Step 8: Calculate x for each y value

For \( y = \sqrt{3.5} \) or \( y = -\sqrt{3.5} \), substitute into \( x = y^2 - 3 \) to get \( x = 0.5 \). For \( y = \sqrt{2} \) or \( y = -\sqrt{2} \), substitute to get \( x = -1 \).

Step 9: Write all solution pairs

Thus, the solutions of the system are: \((0.5, \sqrt{3.5} ), (0.5, -\sqrt{3.5} ), (-1, \sqrt{2}), (-1, -\sqrt{2}) \).

Key Concepts

System of Equations

A system of equations refers to a set of two or more equations that have a common set of unknowns. Solving a system of equations means finding the values of these unknowns that satisfy all equations simultaneously.
In the given exercise, we have two equations involving variables \( x \) and \( y \):

• \( x - y^2 + 3 = 0 \)
• \( 2x^2 + y^2 - 4 = 0 \)

These equations form a system, and our task is to find all pairs \((x, y)\) that satisfy both equations.

To solve this system, we use the elimination method, a technique for combining the equations to eliminate one variable, making it easier to solve for the other. By manipulating the first equation to express \( x \) in terms of \( y \), and then substituting into the second equation, we effectively reduce the system of equations into a single equation to handle.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The solutions to these equations, known as roots, can be found using various methods.
One such method is the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Quadratic equations often represent parabolic graphs and can have two, one, or no real roots depending on the value of the discriminant.

In the given exercise, after simplifying the system, we encounter a quadratic equation in terms of \( z = y^2 \):

• \( 2z^2 - 11z + 14 = 0 \)

This form allows us to employ the quadratic formula to find \( z \), and consequently \( y \), making our way to the solution pairs.

Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). This value provides key information about the nature of the quadratic's roots:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is one real root, also known as a repeated or double root.
• If the discriminant is negative, there are no real roots, but two complex roots instead.

In our example, the discriminant is calculated for the quadratic \( 2z^2 - 11z + 14 = 0 \) as:

• \( (-11)^2 - 4 \cdot 2 \cdot 14 = 9 \)

Since the discriminant is positive, we find two distinct real roots for \( z \), leading to two possible values for \( y^2 \).

These calculations are crucial in understanding the potential solutions and in gaining insights into the behavior of quadratic equations in different contexts.