Question

You and a friend are playing a number guessing game. You ask your friend to think of a positive number, square the number, multiply the result by 2 , and then add 3 . Your friend's final answer is 53 . What was the original number chosen? Justify your answer.

Short answer

The original number was 5.

Step-by-step solution

Rearrange the equation

Rearrange the equation to isolate terms with the unknown on one side. To do this, subtract 3 from both sides of the original equation, giving you the equation \(2x^2 = 50\).

Simplify the equation

Next, simplify the equation by dividing both sides by 2. This results in the equation \(x^2 = 25\).

Solve the simplified equation

Finally, solve this simplified equation by finding the square root of each side. Since the problem specifies that the original number was positive, you can disregard the negative root. Thus, the solution is \(x = \sqrt{25} = 5\). This means the original number was 5.

Key concepts

Mathematical Problem-Solving

Mastering mathematical problem-solving is a crucial skill that allows us to tackle complex challenges systematically. In the context of solving quadratic equations, it involves identifying the given problem, devising a plan, carrying out that plan, and finally reviewing the solution. In the provided exercise, the problem is to reverse-engineer a number from a sequence of operations. The solution plan started with algebraic manipulation and ended with extracting square roots.

Effective problem-solving also requires paying attention to specific details, like the statement that the original number is positive, ensuring we arrive at the correct conclusion. Strategies such as working backwards from the known result are essential, as seen in the step-by-step process to uncover the original number your friend chose.

Algebraic Manipulation

Algebraic manipulation is the art of rearranging and simplifying equations to solve for an unknown variable. It's like a puzzle where we find the pieces (terms) and rearrange them to complete the picture (solution). In our exercise, manipulation began with subtracting 3 from both sides of the equation, which is equivalent to undoing the last operation done to the original number. Subsequent division by 2 reversed the multiplication, leaving us with a pure quadratic equation.

Key Techniques

• Addition or subtraction to isolate terms
• Multiplication or division to simplify coefficients
• Factoring to split complex terms into simpler factors (not explicitly used here, but often essential)

Understanding and applying these techniques is fundamental for any student aiming to excel in algebra.

Square Roots

Square roots are at the heart of solving quadratic equations where the variable is squared. To understand the square root, imagine you have a number which, when multiplied by itself, gives you a certain result. That original number is the square root. In our exercise, after simplifying the equation to \(x^2 = 25\), the next logical step is to find the number that squares to 25.

The square root of 25 is 5, which we represent mathematically as \(\sqrt{25} = 5\). This step is crucial because it embodies the transition from a quadratic equation to a simple, solvable linear situation. It demonstrates an essential operation where precision is key, as squares have two roots: a positive and a negative one. The context of the problem guides us to choose the appropriate root.

Quadratic Formula

The quadratic formula is the silver bullet for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It provides a direct solution where you plug the coefficients into the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) and calculate. Although we did not need to use the quadratic formula in our exercise because it was factorable, knowing this formula is imperative as it can solve equations that are not easily simplified by other means.

In scenarios where algebraic manipulation does not easily yield the roots, or when dealing with coefficients that are not whole numbers, the quadratic formula is our go-to tool. It requires a strong grasp of algebraic manipulation to work through the formula correctly.