Solving Complex Equations with Quadratics: An In-Depth Look at ab 10 and ab 1

Algebra often presents us with complex equations that require intricate solutions. In this article, we will dissect and solve a pair of equations involving the product of two variables, ab 10 and ab 1. We will explore how to approach these equations step-by-step and find the values of the variables a and b.

Understanding the Problem

Let us start with the given equations:

ab 10

ab 1

The equations are similar but have different products on the right-hand sides. At first glance, it may seem contradictory, but we need to approach the problem systematically to find the values of a and b.

Solving the Equations

From the first equation, we can express b in terms of a:

b 10 - a

Now substitute this expression into the second equation:

a(10 - a) 1

Expanding and rearranging the equation gives us:

10a - a^2 1

A standard quadratic equation can be formed:

a^2 - 10a 1 0

Applying the Quadratic Formula

To solve the quadratic equation, we use the quadratic formula:

a frac{-b pm sqrt{b^2 - 4ac}}{2a}

For this specific equation, we have:

a 1

b -10

c 1

Substituting these values into the quadratic formula:

a frac{10 pm sqrt{(-10)^2 - 4 cdot 1 cdot 1}}{2 cdot 1}

Simplifying further:

a frac{10 pm sqrt{100 - 4}}{2}

a frac{10 pm sqrt{96}}{2}

a frac{10 pm 4sqrt{6}}{2}

a 5 pm 2sqrt{6}

Thus, we have two possible values for a:

a 5 2sqrt{6}

a 5 - 2sqrt{6}

Substituting Back to Find b

Substituting these values of a back into the equation b 10 - a:

If a 5 2sqrt{6}, then b 10 - (5 2sqrt{6}) 5 - 2sqrt{6} If a 5 - 2sqrt{6}, then b 10 - (5 - 2sqrt{6}) 5 2sqrt{6}

In both cases, the product ab can be calculated as follows:

ab (5 2sqrt{6})(5 - 2sqrt{6}) 25 - (2sqrt{6})^2 25 - 24 1

Therefore, the value of ab is boxed{1}.

Alternative Approach

Another perspective can be taken when considering the second set of equations. If we have the first equality b 10 - a, and the second equation a frac{1}{n}d, then we substitute into the first:

b 10 - frac{1}{n}d

Thus, ab can be expressed as:

ab frac{(10nd - 1)}{n^2d^2}

Conclusion

This exercise demonstrates the power of algebraic manipulation and the quadratic formula in solving complex equations. Understanding these techniques is essential for advancing in mathematics and many other fields that rely on analytical thinking.