To solve for \( x \) when \( g(x) = 2 \) for the quadratic function \( g(x) = x^2 - 5x \), follow these steps:

### Step 1: Set up the equation
You are given that \( g(x) = 2 \), so:

\[
x^2 - 5x = 2
\]

### Step 2: Move everything to one side of the equation
Subtract 2 from both sides to set the equation equal to zero:

\[
x^2 - 5x - 2 = 0
\]

Now, we have a quadratic equation to solve:

\[
x^2 - 5x - 2 = 0
\]

### Step 3: Solve using the quadratic formula
The quadratic formula is:

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

For the equation \( x^2 - 5x - 2 = 0 \), the coefficients are:
- \( a = 1 \) (the coefficient of \( x^2 \))
- \( b = -5 \) (the coefficient of \( x \))
- \( c = -2 \) (the constant term)

### Step 4: Plug the values into the quadratic formula

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-2)}}{2(1)}
\]

Simplify step by step:

\[
x = \frac{5 \pm \sqrt{25 + 8}}{2}
\]

\[
x = \frac{5 \pm \sqrt{33}}{2}
\]

### Step 5: Simplify the result

The solutions are:

\[
x = \frac{5 + \sqrt{33}}{2} \quad \text{or} \quad x = \frac{5 - \sqrt{33}}{2}
\]

Since \( \sqrt{33} \approx 5.744 \), we can approximate the solutions:

\[
x_1 \approx \frac{5 + 5.744}{2} = \frac{10.744}{2} \approx 5.372
\]
\[
x_2 \approx \frac{5 - 5.744}{2} = \frac{-0.744}{2} \approx -0.372
\]

### Final answer:
The approximate solutions are:
\[
x_1 \approx 5.372, \quad x_2 \approx -0.372
\]

Would you like more details or a follow-up question on this?

Here are five related questions:
1. Solve \( g(x) = 0 \) for \( g(x) = x^2 - 5x \).
2. What are the roots of the equation \( x^2 - 5x - 2 = 0 \)?
3. Find the vertex of the quadratic function \( g(x) = x^2 - 5x \).
4. What is the axis of symmetry for \( g(x) = x^2 - 5x \)?
5. Can you express \( g(x) = x^2 - 5x \) in vertex form?

**Tip:** If the discriminant \( b^2 - 4ac \) is a perfect square, the quadratic can be factored easily. Otherwise, the quadratic formula is the best method to use!

Learn how to solve the quadratic equation g(x) = x^2 - 5x when g(x) = 2 with a step-by-step solution. This example explains how to use the quadratic formula and solve for approximate roots. Suitable 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