The equation is \( y^2 - y = x + 1 \). To find \( y \) in terms of \( x \), we can treat this as a quadratic equation in \( y \) and solve it using the quadratic formula.

The equation is rewritten as:

\[
y^2 - y - (x + 1) = 0
\]

This is a standard quadratic equation of the form \( ay^2 + by + c = 0 \), where:
- \( a = 1 \)
- \( b = -1 \)
- \( c = -(x + 1) \)

Using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values of \( a \), \( b \), and \( c \):

\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-(x + 1))}}{2(1)}
\]

Simplify:

\[
y = \frac{1 \pm \sqrt{1 + 4(x + 1)}}{2}
\]

\[
y = \frac{1 \pm \sqrt{4x + 5}}{2}
\]

So, the solutions for \( y \) are:

\[
y = \frac{1 + \sqrt{4x + 5}}{2} \quad \text{or} \quad y = \frac{1 - \sqrt{4x + 5}}{2}
\]

Would you like further details or clarifications on this?

Here are 5 related questions to extend this concept:

1. How do you determine the domain of the function \( y = \frac{1 \pm \sqrt{4x + 5}}{2} \)?
2. How would you find the derivative of \( y \) with respect to \( x \)?
3. How does the quadratic formula work for equations of different degrees?
4. Can this equation be solved graphically? What does its graph look like?
5. How do complex solutions arise from quadratic equations?

**Tip**: When using the quadratic formula, always check the discriminant (\( b^2 - 4ac \)) to determine the nature of the roots (real or complex).

Discover how to solve the quadratic equation y^2 - y = x + 1 using the quadratic formula. This step-by-step guide explains how to handle quadratic equations, including extracting solutions for y. Ideal for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation