Problem 15 - Algebra A Combined Function

Question

Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} y=2 x^{2}+1 \\ x+y=-1 \end{array}\right. $$

Step-by-Step Solution

Answer

No real solutions.

Step 1: Substitute y in the second equation

Given the first equation is \( y = 2x^2 + 1 \). Substitute this expression for \( y \) in the second equation \( x + y = -1 \). This gives us the equation \( x + (2x^2 + 1) = -1 \).

Step 2: Simplify the resulting equation

Now simplify the equation \( x + 2x^2 + 1 = -1 \) by combining like terms. Subtract 1 from -1 which gives: \( 2x^2 + x + 1 = -1 \). Then subtract 1 from both sides to have \( 2x^2 + x + 1 - 1 = -1 - 1 \). Simplify this to get \( 2x^2 + x = -2 \).

Step 3: Set equation to zero

Add 2 to both sides of the equation \( 2x^2 + x = -2 \) to set it to zero: \( 2x^2 + x + 2 = 0 \).

Step 4: Solve the quadratic equation

To find \( x \), solve the quadratic equation \( 2x^2 + x + 2 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=2 \), \( b=1 \), \( c=2 \). Calculate the discriminant \( b^2 - 4ac = 1^2 - 4(2)(2) = 1 - 16 = -15 \). Since the discriminant is negative, there are no real roots.

Key Concepts

Quadratic Equations

Quadratic equations form a cornerstone in algebra, representing polynomial equations of the second degree. The standard form of a quadratic equation is given by \[ ax^2 + bx + c = 0 \] where \( a, b, \) and \( c \) are constants, with \( a eq 0 \). This equation graphs as a parabola, which might open upwards or downwards depending on the sign of \( a \). Quadratic equations can have either:

• Two real solutions
• One real solution (in case of a double root)
• No real solutions

These possibilities derive from the nature of the parabola in relation to the x-axis. Quadratic equations can be addressed using various methods: factoring, completing the square, or employing the quadratic formula. In our given exercise, the quadratic formula helps us solve the quadratic equation derived from the original nonlinear system.

Substitution Method

The substitution method is a technique used to solve systems of equations, especially useful for systems involving a mix of linear and nonlinear equations. Here’s how it works:

• Step 1: Solve one of the equations for one variable in terms of the other variable. This equation might be either linear or nonlinear.
• Step 2: Substitute this expression into the other equation. This step changes the original system into a single equation with one variable, simplifying the solution process.

In the provided solution, the first equation \( y = 2x^2 + 1 \) was already solved for \( y \). This expression is then substituted into the second equation \( x + y = -1 \) to eliminate \( y \), thus forming a quadratic equation in terms of \( x \). This method greatly reduces complexity and is best used when one of the equations allows easy isolation of a variable.

Discriminant

The discriminant is a critical part of the quadratic formula, helping to determine the number of real solutions of a quadratic equation. It is represented by:\[ D = b^2 - 4ac \]Here's what the discriminant reveals:

• If \( D > 0 \): There are two distinct real solutions.
• If \( D = 0 \): There is exactly one real solution (a repeated root).
• If \( D < 0 \): There are no real solutions, only complex numbers.

In the given exercise, once we formed and simplified the quadratic equation \( 2x^2 + x + 2 = 0 \), calculating the discriminant \( D \) revealed a negative value:\ 15. This negative discriminant immediately informs us that the quadratic equation has no real solutions, implying that the original system of nonlinear equations does not intersect at any real-numbered point for \( x \) and \( y \). This insight into the nature of solutions further highlights the importance of analyzing the discriminant in any quadratic equation scenario.