Problem 122 - College Algebra

Question

Will help you prepare for the material covered in the next section. $$ \text { Solve for } y: \quad x-y^{2}-1, y \geq 0 $$

Step-by-Step Solution

Answer

The solution for \(y\) is \(y = \sqrt{x - 1}\), valid for \(x \geq 1\).

Step 1: Rearrange Equation

First, rearrange the given equation \(x = y^{2} + 1\) to get \(y^{2} = x - 1\).

Step 2: Solve for y

Take the square root of both sides of the equation \(y^{2} = x - 1\). However remember that usually, with square roots, two solutions are obtained, a positive and negative root. Therefore \(y = \sqrt{x - 1}\) or \(y = -\sqrt{x-1}\).

Step 3: Apply Non-negative Condition

But the condition \(y >= 0\), implies that \(y\) cannot be negative. Therefore, we discard the negative root. We're now left with the solution \(y = \sqrt{x - 1}\). However, this solution is valid as long as \(x - 1 >= 0\), i.e., \(x >= 1\). So the final solution to the equation is \(y = \sqrt{x - 1}\), \(x>=1\).

Key Concepts

Square roots

Square roots are an essential operation in algebra and dealing with quadratic equations. They provide a method to "undo" squaring, which is why they appear frequently when solving such equations. When you take the square root of a number, you essentially find a value that, when multiplied by itself, equals the original number.

For instance, if you have the equation \( y^2 = x - 1 \), taking the square root of both sides allows you to find \( y \). This results in two possible roots: \( y = \sqrt{x - 1} \) and \( y = -\sqrt{x - 1} \). These are called the principal square root (the positive root) and the negative root.

• **Principal Square Root:** This is what we commonly refer to when we say 'square root.' In mathematical notation, it is represented as \( \sqrt{x} \). It is always non-negative.
• **Negative Root:** This is the opposite value of the principal square root and is not often considered unless specified.

When solving equations involving square roots, it is crucial to identify which root is required by the problem's conditions, like non-negativity, which we'll explore later.

Equation rearrangement

Rearranging an equation is a fundamental skill in algebra that forms the basis of solving equations. The idea is to manipulate the equation by performing legal operations to isolate the variable of interest on one side of the equation. This makes it easier to find its value.

Consider the equation given in the exercise: \( x = y^2 + 1 \). Our goal is to solve for \( y \), so we need to rearrange the equation to express \( y \) in terms of other variables or constants. This involves:

• Subtracting 1 from both sides to move constant terms away from the \( y^2 \) term:
• Resulting in: \( y^2 = x - 1 \).

This square term isolation prepares you for the next step, which involves using square roots to solve for \( y \). Effective rearrangement helps in seeing the structure of the equation more clearly and sets the stage for subsequent operations.

Non-negative condition

A non-negative condition in mathematics specifies that a variable must be greater than or equal to zero. This condition is crucial when dealing with square roots, as square root outputs by default are assumed non-negative.

In the solution for our exercise, after finding \( y = \sqrt{x - 1} \) and \( y = -\sqrt{x-1} \), the non-negative condition \( y \geq 0 \) influences the choice of solution. Given the condition, only \( y = \sqrt{x - 1} \) is a valid solution because it is non-negative.

• This condition ensures that the solution aligns with real-world scenarios where negative values might not make sense or could be undefined.
• For instance, if \( y \) represented a distance or a physical quantity that can't be negative, this condition is essential.

Additionally, the derived condition \( x \geq 1 \) from \( x - 1 \geq 0 \) follows because \( \sqrt{x - 1} \) must be defined. This ensures the expression under the square root is non-negative, ensuring a real, usable solution.