Question

Divide 10 into two parts such that their product is \(13+\) \(\sqrt{128}\)

Short answer

Question: Divide the number 10 into two parts such that their product is equal to \(13+\sqrt{128}\). Find the two parts. Answer: The two parts are approximately \(\left(\frac{10 + \sqrt{48 - 32\sqrt{2}}}{2}, 10 - \frac{10 + \sqrt{48 - 32\sqrt{2}}}{2}\right)\) and \(\left(\frac{10 - \sqrt{48 - 32\sqrt{2}}}{2}, 10 - \frac{10 - \sqrt{48 - 32\sqrt{2}}}{2}\right)\).

Step-by-step solution

Define the unknowns

Let the two parts be x and y. We have two conditions: the sum of the parts is 10, and their product is \(13+\sqrt{128}\). So we can write: x + y = 10 x * y = 13 + sqrt{128}

Expression for one part as the other

We can rewrite the first equation to find y in terms of x: y = 10 - x

Substitute in the second equation

Now, we will substitute the expression for y into the second equation: x * (10 - x) = 13 + sqrt{128}

Solve the quadratic equation for x

We need to solve the quadratic equation for x: x^2 - 10x + 13 + sqrt{128} = 0 Note that the coefficient for x^2 is positive, so it is an upward-opening parabola. Since we are asked to divide 10 into two parts, naturally there will be two solutions for x.

Use the quadratic formula to solve for x

We can solve the quadratic equation for x using the quadratic formula: x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In our case, a = 1, b = -10, and c = 13 + sqrt{128}. Plugging these values in the quadratic formula, we get: x = \(\frac{10 \pm \sqrt{(-10)^2 - 4(1)(13 + \sqrt{128})}}{2(1)}\) x = \(\frac{10 \pm \sqrt{100 - 52 - 4\sqrt{128}}}{2}\) x = \(\frac{10 \pm \sqrt{48 - 4\sqrt{128}}}{2}\)

Simplify the expression for x

Simplify the expression under the square root: 48 - 4*sqrt{128} = 48 - 4*sqrt{64*2} = 48 - 4*8*sqrt{2} = 48 - 32*sqrt{2} Therefore, x = \(\frac{10 \pm \sqrt{48 - 32\sqrt{2}}}{2}\)

Calculate the two parts

Now, let's calculate the two parts using the two solutions for x: x1 = \(\frac{10 + \sqrt{48 - 32\sqrt{2}}}{2}\) x2 = \(\frac{10 - \sqrt{48 - 32\sqrt{2}}}{2}\) Then, plug these solutions into the expression for y: y1 = 10 - x1 = \(10 - \frac{10 + \sqrt{48 - 32\sqrt{2}}}{2}\) y2 = 10 - x2 = \(10 - \frac{10 - \sqrt{48 - 32\sqrt{2}}}{2}\) Thus, the two parts are: Part1: x1, y1 = \(\left(\frac{10 + \sqrt{48 - 32\sqrt{2}}}{2}, 10 - \frac{10 + \sqrt{48 - 32\sqrt{2}}}{2}\right)\) Part2: x2, y2 = \(\left(\frac{10 - \sqrt{48 - 32\sqrt{2}}}{2}, 10 - \frac{10 - \sqrt{48 - 32\sqrt{2}}}{2}\right)\)

Key concepts

Solving Quadratic Equations

Understanding how to solve quadratic equations is fundamental in algebra. A quadratic equation typically takes the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. The solutions to this type of equation represent the points where the graph of the quadratic function intersects the x-axis.

In the exercise provided, we translate a word problem into a quadratic equation by defining the variables and setting up equations based on the given conditions. After expressing one variable in terms of the other and substituting, we reach a standard quadratic equation. Solving it reveals the parts we're looking for. The process is methodical and applies to diverse problems requiring quadratic solutions.

Algebraic Expressions

Algebraic expressions are the building blocks of algebra. They consist of numbers, variables, and arithmetic operations. In the context of our exercise, \( x + y = 10 \) and \( x * y = 13 + \sqrt{128} \) are two algebraic expressions that represent the sum and product of two parts we aim to find.

The expression's manipulation, such as isolating one variable and substituting into another equation, demonstrates the power of algebraic expressions in formulating and solving mathematical problems. By understanding how to manipulate these expressions, students can solve complex problems step-by-step.

Mathematical Problem Solving

Mathematical problem solving is more than just computation; it involves understanding the problem, devising a plan, carrying out the plan, and then looking back to check the results. The given exercise showcases this process: we start by defining variables, create equations based on the problem conditions, manipulate these equations, apply mathematical methods, and reach a solution that we can interpret in the context of the original problem.

It's essential that students learn to transition from a written problem to an algebraic representation, and from there to employ various problem-solving techniques, such as the quadratic formula, to find solutions.

Quadratic Formula

The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is used to find the solutions of a quadratic equation. It is derived from completing the square of a general quadratic equation and provides a direct way to find the roots of any quadratic equation.

Within our exercise, after arriving at a quadratic equation, we apply the quadratic formula. This step is crucial; it allows us to find the exact numerical values of the two parts that satisfy both the sum and the product conditions. By using the quadratic formula, we can solve what might otherwise be a complex equation through basic arithmetic operations.