Problem 9 - A History of Mathematics: An Introduction

Question

Divide 10 into two parts such that if one squares the first, subtracts it from 97, and takes its square root, then squares the second, subtracts it from 100 , and takes its square root, the sum of the two roots is 17. (This problem is also from the work of Antonio de' Mazzinghi. Mazzinghi set the parts \(u, v\) equal to \(5+x\) and \(5-x\), respectively, and derived an equation in \(x .\) )

Step-by-Step Solution

Answer

Question: Divide 10 into two parts such that the sum of the square roots of (97 minus the square of the first part) and (100 minus the square of the second part) is 17. Answer: The two parts that divide 10 and satisfy the given conditions are \(u = \frac{27}{4}\) and \(v = \frac{7}{4}\).

Step 1: Define u and v

Let u and v be the two parts of 10. According to Mazzinghi, set u = 5+x and v = 5-x, where x is a variable we need to find.

Step 2: Write the equations for the square roots

Given the conditions, we have the following equations for the square roots: \(\text{Square root 1}=\sqrt{97-u^2}\) \(\text{Square root 2}=\sqrt{100-v^2}\)

Step 3: Write the equation for the sum of the square roots

We are given that the sum of the square roots is 17. So we can write: \(\sqrt{97-u^2} + \sqrt{100-v^2} = 17\)

Step 4: Substitute u and v with 5+x and 5-x

Now, replace u and v with the respective expressions from Step 1. We get: \(\sqrt{97-(5+x)^2}+\sqrt{100-(5-x)^2} = 17\)

Step 5: Simplify the equation

In this step, we'll simplify the equation by first expanding the squares and then collecting terms: \(\sqrt{97- (25+10x+x^2)}+\sqrt{100-(25-10x+x^2)} = 17\) \(\sqrt{-x^2+10x+72}+\sqrt{-x^2+10x+75}=17\)

Step 6: Solve the equation for x

It's a bit difficult to solve this nonlinear equation directly, but we can make use of the initial assumption that u+v=10. Notice that the term under each square root is the square of the difference of u or v with some constant. So if we add the two square roots using the variables u and v, we have: \(\sqrt{u^2-2u\cdot10+72}+\sqrt{v^2-2v\cdot10+75}=17\) \(\sqrt{(u-10)^2} + \sqrt{(v-10)^2}= 17\) Since u and v are both positive (which can be assumed since we are asked for "dividing 10"), it follows that u - 10 and v - 10 are both negative. Therefore, the expressions under the square roots are the squares of their absolute values: \(|u-10| + |v-10| = 17\) Now, we can use our initial assumption that u+v=10: \(|(5+x)-10| + |(5-x)-10| = 17\) \(|-5+x| + |-5+x| = 17\) Without loss of generality, let -5+x be a positive value: \(x+5 = |-5+x|\) Since we are assuming x is positive, the equation becomes: \(2x+10 = 17\) \(x = \frac{1}{2}(17-10)\) \(x = \frac{7}{2}\)

Step 7: Find u and v

We have found the value of x, now we can substitute it back into the expressions for u and v from Step 1: \(u = 5+\frac{7}{2}=5+\frac{14}{4}=\frac{27}{4}\) \(v = 5-\frac{7}{2}=5-\frac{14}{4}=\frac{7}{4}\)

Step 8: Conclusion

The two parts that divide 10 and satisfy the given conditions are \(u = \frac{27}{4}\) and \(v = \frac{7}{4}\).

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. These equations can have two solutions, one solution, or no real solutions, depending on the discriminant \( b^2 - 4ac \). This concept is fundamental in algebra and can often be solved using various methods such as factoring, completing the square, or using the quadratic formula:

• The quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Factoring: Requires the equation to be written in a form where it can be expressed as a product of two binomials.
• Completing the Square: Involves rearranging the equation so that one side becomes a perfect square trinomial.

In this particular exercise, we encounter a quadratic situation while handling the expressions under the square roots, as the relationship \( u = 5+x \) and \( v = 5-x \) leads us to squared expressions. Understanding how to recognize and handle equations in quadratic form is crucial here.

Variable Substitution

Variable substitution is a technique used to simplify complex equations by replacing variables with new expressions. This method makes solving equations easier and more understandable. In our exercise, we used the substitution \( u = 5 + x \) and \( v = 5 - x \). This effectively transforms the problem into one that is algebraically more tractable.

By substituting \( x \), we transition the given conditions into the realm of quadratic expressions, enabling further manipulation and simplification. This approach allows us to focus on the essential relationships and avoid getting bogged down by more complicated arithmetic expressions. Substitution is particularly useful:

• For reducing the number of variables.
• To transform a problem into a simpler mathematical form.
• When dealing with symmetry in equations.

This method is not limited just to algebraic problems but is widely employed in calculus and other mathematical applications.

Square Roots

Square roots play a significant role in this exercise. The square root of a number \( y \) is a value \( x \) such that \( x^2 = y \). In algebra, square roots are often used to solve quadratic equations and are important in simplifying expressions. The initial conditions of our problem involve the sums of two square roots:

• First root: \( \sqrt{97-u^2} \)
• Second root: \( \sqrt{100-v^2} \)
• The given sum: \( \sqrt{97-u^2} + \sqrt{100-v^2} = 17 \)

Handling square roots often requires attention to detail, especially when they appear under complex expressions. Simplifying or solving square root equations is possible by raising both sides to the power of two, which removes the root symbol. However, this must be done carefully to avoid introducing extraneous solutions or errors.

Understanding properties of square roots, such as \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \) and \( \sqrt{a^2} = |a| \), is essential for mastering problems like these. When paired with quadratic expressions or variable substitutions, square roots often require a deeper analysis to maintain the correct mathematical relationships in an equation.