Problem 19 - Algebra and Trigonometry

Question

\(19-32\) . 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

There are no real solutions to the system of equations.

Step 1: Rewrite and Subtract the Equations

Given the system of equations: \[ \begin{align*} y + x^2 &= 4x, \ y + 4x &= 16. \end{align*} \] Subtract the second equation from the first to eliminate \(y\). This results in: \[ (y + x^2) - (y + 4x) = 4x - 16. \] Simplifying, we obtain \(x^2 - 4x = -16\).

Step 2: Rearrange the Equation

Rearrange the equation from Step 1 to form a standard quadratic equation: \[ x^2 - 4x + 16 = 0. \] However, notice earlier we simplified as \(x^2 = 4x - 16\). Re-evaluate checking simplifications.

Step 3: Solve the Quadratic Equation

Solve the simplified quadratic equation \(x^2 - 4x = -16\) by adding 16 to both sides to give: \[ x^2 - 4x + 16 = 0. \] Determine if this equation has real solutions using the discriminant \(b^2 - 4ac\): \[ (-4)^2 - 4(1)(16) = 16 - 64 = -48. \] Since the discriminant is negative, there are no real solutions to the quadratic equation.

Step 4: Evaluate for \(y\) (Theoretical)

Although Step 3 indicates no real solutions, if theoretically solved for \(x\), substitute back to original: \[ y = 4x - x^2. \] Given real \(x\) this does not occur.

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation where the highest exponent of the variable is two. In the standard form, a quadratic equation is expressed as:\[ ax^2 + bx + c = 0 \]where:

• \( a \), \( b \), and \( c \) are constants with \( a eq 0 \).
• \( x \) is the variable.

Quadratics can graph as parabolas, opening upwards if \( a > 0 \) and downwards if \( a < 0 \).
To solve quadratic equations, you can use several methods such as factoring, completing the square, or using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Understanding how to arrange an equation into this standard form is crucial before applying these solving methods.
In our exercise, rearranging was necessary to align our equation to the quadratic form, which allows further analysis.

Discriminant

The discriminant is a specific component of the quadratic formula, represented as \( b^2 - 4ac \). It offers crucial information about the nature of the roots of a quadratic equation:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one true real solution (also described as a repeated or double root).
• If \( b^2 - 4ac < 0 \), there are no real solutions, but two complex solutions instead.

Using the discriminant is a quick way to ascertain whether real solutions exist without fully solving the equation.
In the given exercise, the discriminant calculation was performed as:\[ (-4)^2 - 4(1)(16) = 16 - 64 = -48 \]The negative result here indicates that the quadratic equation formed from our system has no real solutions.

Real Solutions

Real solutions in a system of equations refer to the values for the variable(s) that satisfy the equations simultaneously with real numbers. For quadratic equations, the real solutions correspond to the points where the parabola intersects the x-axis.
When dealing with a system like the one in the exercise, finding real solutions entails exploring where the solutions satisfy all involved equations.
Given that the discriminant in this system's quadratic equation is less than zero, it confirms that there are no real values of \( x \) that satisfy the entire system.
You may theoretically find complex solutions, but for this exercise and in real-world scenarios, we're often more interested in real solutions. If none exist due to the discriminant's nature, it implies the system has no real intersection points, as noted in our analysis.