Problem 23 - Algebra and Trigonometry

Question

$$ \left\\{\begin{aligned} x-y &=4 \\ x y &=12 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Solutions are \((6, 2)\) and \((-2, -6)\).

Step 1: Express one variable in terms of the other

From the first equation, we have \(x - y = 4\). We can express \(x\) in terms of \(y\): \(x = y + 4\).

Step 2: Substitute into the second equation

Substitute \(x = y + 4\) into the second equation \(xy = 12\), so \((y + 4)y = 12\).

Step 3: Expand and form a quadratic equation

Expanding the equation \((y + 4)y = 12\), we get \(y^2 + 4y = 12\). Moving 12 to the left side, we obtain the quadratic equation: \(y^2 + 4y - 12 = 0\).

Step 4: Solve the quadratic equation

To solve \(y^2 + 4y - 12 = 0\), use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1, b = 4, c = -12\). Substitute these values into the formula.

Step 5: Calculate the discriminant and solutions for y

Calculate the discriminant: \(D = b^2 - 4ac = 16 + 48 = 64\). The solutions for \(y\) are \(y = \frac{-4 \pm 8}{2}\). Hence, \(y = 2\) or \(y = -6\).

Step 6: Find corresponding x values

For \(y = 2\), substitute back into \(x = y + 4\), giving \(x = 6\). For \(y = -6\), \(x = -2\).

Step 7: Verify solutions

Check \(x = 6, y = 2\) in both original equations: \(6 - 2 = 4\) and \(6 \times 2 = 12\). Check \(x = -2, y = -6\): \(-2 + 6 = 4\) and \((-6) \times (-2) = 12\). Both pairs satisfy the equations.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2, typically expressed in the form \( ax^2 + bx + c = 0 \). It is characterized by its highest power being a square, which is why it's called 'quadratic,' from the Latin word 'quadratus' meaning 'square.'
In the context of solving simultaneous equations, like in our original exercise, forming a quadratic equation involves combining the results of substituting one equation into another.

• The coefficients \(a\), \(b\), and \(c\) are essential as they dictate the shape and position of the parabola described by the equation.
• The solutions to these equations, also known as the roots, can be real or complex depending on the discriminant, a concept we will explore later.

Quadratic equations often appear in physics and engineering problems and are essential in finding relationships that describe parabolic trajectories, among other things. Understanding how to convert problems into quadratic equations and solve them is crucial.

Substitution Method

The substitution method is a technique for solving simultaneous equations. It involves expressing one variable with the help of another equation and substituting this expression into the other equation.
This process effectively reduces two equations with two unknowns into a single equation with one unknown.

• It is especially useful when one equation is linear and the other is not, as seen in the original exercise.
• By substituting, you simplify the problem, often turning it into a more familiar type, such as a quadratic equation.

In practice, it requires careful attention to detail, as small errors in substitution could lead to incorrect solutions. The key is to isolate one variable, as shown in Step 1 of the original solution, and use it to solve the quadratic formed.

Discriminant

The discriminant is a component of the quadratic formula used to determine the nature of the roots of a quadratic equation. It is derived from the coefficients of the equation \( ax^2 + bx + c = 0 \) as \( D = b^2 - 4ac \).

• If \( D > 0 \), the quadratic equation has two distinct real roots.
• If \( D = 0 \), there is exactly one real root, also called a repeated or double root.
• If \( D < 0 \), the equation has two complex roots that are not real numbers.

For our equation \( y^2 + 4y - 12 = 0 \), the discriminant is 64, which indicates there are two distinct real solutions. Understanding and calculating the discriminant is important for predicting the types of solutions you'll encounter in a quadratic equation, allowing you to approach complex problems with confidence.