Problem 19 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l} y+x^{2}=4 x \\ y+4 x=16 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = 4 \) and \( y = 0 \).

Step 1: Solve the Second Equation for y

We start by solving the second equation for \( y \). The equation is \( y + 4x = 16 \). Subtract \( 4x \) from both sides to isolate \( y \):\[ y = 16 - 4x \].

Step 2: Substitute for y in the First Equation

Now take the expression for \( y \) from Step 1 and substitute it into the first equation \( y + x^2 = 4x \). This gives us:\[ (16 - 4x) + x^2 = 4x \].

Step 3: Simplify the Equation

Distribute and simplify the equation from Step 2:\[ x^2 - 4x + 16 = 4x \].Subtract \( 4x \) from both sides:\[ x^2 - 8x + 16 = 0 \].

Step 4: Solve the Quadratic Equation

The equation is a perfect square trinomial. Recognize and factor it as:\[ (x - 4)^2 = 0 \].This gives \( x = 4 \) since the only solution to a perfect square is when it's zero.

Step 5: Find the Corresponding y Value

Substitute \( x = 4 \) back into the equation \( y = 16 - 4x \) from Step 1:\[ y = 16 - 4(4) \].Calculate to find \( y = 16 - 16 = 0 \).

Step 6: Verify the Solution

Substitute \( x = 4 \) and \( y = 0 \) back into the original equations to verify:- For the first equation, \( y + x^2 = 4x \) becomes \( 0 + 16 = 16 \), which holds true.- For the second equation, \( y + 4x = 16 \) becomes \( 0 + 16 = 16 \), which also holds true.Both equations are satisfied, confirming the solution.

Key Concepts

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. In systems of equations like the one in the exercise, quadratic equations often need solving to find values of \( x \). The standard methods to solve them include:

• Factoring
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

Quadratic equations can have two solutions, one solution or none, depending on the discriminant\( (b^2 - 4ac) \). A positive discriminant means two real solutions, zero means one solution (a repeated root), and negative means no real solutions.

Substitution Method

The substitution method is a technique used to solve systems of equations, especially when one of the equations can easily be solved for one variable. This method involves:

• Solving one of the equations for one variable in terms of the other. For instance, solving \( y + 4x = 16 \) for \( y \) gives \( y = 16 - 4x \).
• Substituting this expression into the other equation. In the original task, we substitute \( y = 16 - 4x \) into \( y + x^2 = 4x \), leading to a single equation with one variable.
• Simplifying and solving the resulting equation.

This method simplifies the system to an equation that is often easier to solve, reducing the complexity of handling multiple variables at once. Once you solve for one variable, you can substitute back to find the other.

Perfect Square Trinomial

A perfect square trinomial is a type of quadratic expression that can be pretty easily factored as the square of a binomial. It has the form \( (ax + b)^2 = a^2x^2 + 2abx + b^2 \). In the given exercise, the quadratic equation \( x^2 - 8x + 16 = 0 \) can be recognized as a perfect square trinomial.To identify a perfect square trinomial:

• Check if the first and last terms are perfect squares.
• Ensure the middle term is twice the product of the square roots of the first and last terms.

For the equation \( x^2 - 8x + 16 \), note that \( x^2 \) is \( (x)^2 \) and 16 is \( (4)^2 \). The middle term \(-8x\) is twice the product of \( x \) and \(-4\), confirming it as a perfect square. The equation factors into \( (x - 4)^2 = 0 \), hence the solution is \( x = 4 \). This kind of simplification ensures the solution is quickly found by setting the binomial expression to zero.