Solving for Numbers Given Their Product and Difference

Today, we will explore a classic problem in mathematics involving two numbers where we know their product and difference. Specifically, we are given that the product of two numbers is 4, and their difference is 3. Let's denote these two numbers as x and y.

Setting Up Equations

We can set up the following equations based on the given conditions:

Equation 1: The product of the two numbers: x * y 4 Equation 2: The difference of the two numbers: x - y 3

Solving the Equations

From the second equation, we can express x in terms of y:
x y 3

Next, we substitute this expression for x into the first equation:

(y 3) * y 4

Expanding this equation, we get:

y^2 3y 4

Rearranging the equation to standard form, we obtain:

y^2 3y - 4 0

This is a quadratic equation that we can solve by factoring:

(y 4)(y - 1) 0

Setting each factor to zero gives us the possible solutions for y: y 4 0 → y -4 y - 1 0 → y 1

Now, we can find the corresponding values of x for each value of y:

For y -4, the corresponding value of x is: x -4 3 -1 So one pair of numbers is -1, -4. For y 1, the corresponding value of x is: x 1 3 4 So the other pair of numbers is 4, 1.

Thus, the two pairs of numbers that satisfy the conditions are: -1, -4 and 4, 1.

Verification

To verify these solutions, we can check the product and difference for each pair of numbers:

For the pair -1, -4: Product: -1 * -4 4 Difference: -1 - (-4) 3

For the pair 4, 1: Product: 4 * 1 4 Difference: 4 - 1 3

Both pairs satisfy the given conditions.

Therefore, the two pairs of numbers that meet the specified conditions are: -1, -4 and 4, 1.

Potential Errors in Other Solutions

Note that some of the provided solutions contain errors. For example:

The solution claiming 4 times; 1 4 and 4 - 1 3 is correct but does not provide the part where it confirms the pair (-1, -4). Another solution incorrectly suggests x 4, y 0, which does not satisfy the conditions due to the difference being 4, not 3, and the product being 0, not 4. The last example concludes the pairs to be 4 times; 0 0 which does not meet the conditions since the product is 4, not 0.

By following the steps above, we can accurately solve for the desired numbers given their product and difference.