Let's solve the equation:

\[
3x^2 + 2x - 4 = (2x - 1)^2
\]

### Step 1: Expand the right-hand side
First, expand \((2x - 1)^2\):

\[
(2x - 1)^2 = (2x - 1)(2x - 1) = 4x^2 - 4x + 1
\]

Now the equation becomes:

\[
3x^2 + 2x - 4 = 4x^2 - 4x + 1
\]

### Step 2: Move all terms to one side
Subtract \(4x^2 - 4x + 1\) from both sides:

\[
3x^2 + 2x - 4 - (4x^2 - 4x + 1) = 0
\]

Simplifying this:

\[
3x^2 + 2x - 4 - 4x^2 + 4x - 1 = 0
\]

\[
(3x^2 - 4x^2) + (2x + 4x) + (-4 - 1) = 0
\]

\[
-x^2 + 6x - 5 = 0
\]

### Step 3: Solve the quadratic equation
Now we have the quadratic equation:

\[
-x^2 + 6x - 5 = 0
\]

To make things simpler, multiply through by \(-1\) to get:

\[
x^2 - 6x + 5 = 0
\]

### Step 4: Factor the quadratic equation
Now, factor the quadratic equation:

\[
x^2 - 6x + 5 = (x - 5)(x - 1) = 0
\]

### Step 5: Solve for \(x\)
Set each factor equal to zero:

\[
x - 5 = 0 \quad \text{or} \quad x - 1 = 0
\]

So, \(x = 5\) or \(x = 1\).

### Final answer:
The solutions are \(x = 5\) and \(x = 1\).

Would you like further details or explanations on any step?

Here are five related questions for you:
1. How do you expand binomials like \((2x - 1)^2\)?
2. Can you solve other types of quadratic equations?
3. What methods can be used if a quadratic equation can't be factored easily?
4. How would the solution change if the equation involved cube terms instead of squares?
5. Can you solve the equation using the quadratic formula instead of factoring?

**Tip:** Always check your solutions by plugging them back into the original equation to ensure they are correct!

Learn how to solve the quadratic equation 3x² + 2x - 4 = (2x - 1)² with detailed, step-by-step instructions. This problem covers core algebraic concepts, including expanding binomials, factoring, and solving quadratic equations. Perfect for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation