Question

The product of two numbers is 4 and the sum of their squares is 8 . Find the numbers.

Short answer

The numbers are 2 and 2 or -2 and -2.

Step-by-step solution

Define Variables

Let the two numbers be denoted by variables, say, x and y. Thus, we have two equations based on the problem statement: 1. The product of the numbers: \[ x \times y = 4 \] 2. The sum of their squares: \[ x^2 + y^2 = 8 \]

Express y in Terms of x

From the product equation, express one variable in terms of the other. Using \( x \times y = 4 \): \[ y = \frac{4}{x} \]

Substitute into the Sum of Squares Equation

Substitute \( y = \frac{4}{x} \) into the sum of squares equation: \[ x^2 + \left( \frac{4}{x} \right)^2 = 8 \]Simplify the equation: \[ x^2 + \frac{16}{x^2} = 8 \]

Multiply by x^2 to Eliminate Denominator

Multiply the entire equation by \( x^2 \) to eliminate the fraction: \[ x^4 + 16 = 8x^2 \]

Form a Quadratic Equation

Rearrange the equation to form a standard quadratic equation: \[ x^4 - 8x^2 + 16 = 0 \]Let \( z = x^2 \). The equation then becomes: \[ z^2 - 8z + 16 = 0 \]

Solve the Quadratic Equation

Solve the quadratic equation \( z^2 - 8z + 16 = 0 \). Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find: \[ z = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} = 4 \]

Solve for x and y

Since \( z = x^2 = 4 \), solve for x to get \( x = \pm 2 \). Substitute these values back into the product equation to find y:1. If \( x = 2 \), then \( y = \frac{4}{2} = 2 \).2. If \( x = -2 \), then \( y = \frac{4}{-2} = -2 \).

Key concepts

system of equations

In this exercise, we dealt with a system of equations. A system of equations is a set of two or more equations with the same variables. In this case, we used two equations: the product of two numbers and the sum of their squares. That's what forms our system. By solving these equations together, we can find values for the variables that satisfy all equations in the system.

The first equation we have is:

• \[ x \times y = 4 \]
• The second equation is: \[ x^2 + y^2 = 8 \]

Using both these equations together allows us to pinpoint exactly what the numbers are.

substitution method

Here, we use the substitution method to solve our system of equations. The substitution method involves solving one of the equations for one variable, and then substituting that expression into the other equation.

For example, from \[ x \times y = 4 \], we can isolate one of the variables (let’s pick y) and solve for it in terms of the other variable:

• \[ y = \frac{4}{x} \]

This gives us a single variable to work with in the other equation. We substitute this expression for y into the second equation. Now, our second equation looks like:

\[ x^2 + \frac{16}{x^2} = 8 \]

This simplification helps us move forward with solving our system.

quadratic formula

When our equations are simplified, we often end up with a quadratic equation. The quadratic formula is a powerful tool for solving quadratic equations of the form \[ax^2 + bx + c = 0 \]. The formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] gives us the solution to these equations.

In our example here, we rearranged the equation after substitution and simplification to \[ x^4 - 8x^2 + 16 = 0 \]. For ease of solving, we let \[ z = x^2 \], converting it into a quadratic \[z^2 - 8z + 16 = 0 \]. Using the quadratic formula, we solved:

• \ \[ z = \frac{8 \pm \sqrt{64 - 64}}{2} = 4 \]

This step provided us values for \[ x \] on back-substitution of our solved \[ z = x^2 \].

algebraic manipulation

The main backbone of solving this problem was algebraic manipulation. It refers to rearranging and simplifying expressions to solve for the unknowns. Every step involves some level of this manipulation to drive the equation to a solvable state.

For instance, turning \[ y = \frac{4}{x} \] and substituting into \[ x^2 + y^2 = 8 \] is algebraic manipulation. Additionally, multiplying by \[ x^2 \] to clear the fraction and then rearranging:

• \[ x^4 + 16 = 8x^2 \] \[ x^4 - 8x^2 + 16 = 0 \]

All these steps involve algebraic rules to isolate and solve the variables step-by-step, leading us to the final solution for \[ x \] and \[ y \]. Understanding how to manipulate these expressions is key in solving more complex equations efficiently.