Question

A student solved a quadratic equation as shown here:
2x^2+5x=12
x(2x+5)=12
x=12 or 2x+5=12
x=12 or 2x+5-5=12-5
x=12 or 2x=7
x=12 or x=7/2
x=7/2,12

Explain the errors in the process. Then correct the errors, and solve the equation.

Answer 1

The student made several errors in the process of solving the quadratic equation \(2x^2+5x=12\). Let's break it down:

1. **Incorrect Factoring Step**:
The student starts with the equation \(2x^2 + 5x = 12\) and changes it to \(x(2x + 5) = 12\). This is incorrect because you cannot factor \(2x^2 + 5x\) this way.

2. **Incorrect Case of Setting Parts Equal**:
The student then sets the equation \(x(2x + 5) = 12\) as \(x = 12\) or \(2x + 5 = 12\). However, the first step \(x(2x + 5) = 12\) does not directly lead to \(x = 12\). Instead, one must rearrange the equation correctly in order to find the roots.

### Correct Way to Solve the Equation

1. First, move all terms to one side to set the equation to zero:

\[
2x^2 + 5x - 12 = 0
\]

2. Now, apply the quadratic formula:

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

where \(a = 2\), \(b = 5\), and \(c = -12\).

3. Calculate the discriminant:

\[
b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-12) = 25 + 96 = 121
\]

4. Now, substitute back into the quadratic formula:

\[
x = \frac{-5 \pm \sqrt{121}}{2 \cdot 2}
\]

\[
x = \frac{-5 \pm 11}{4}
\]

5. Solve for the two possible values:

\[
x_1 = \frac{6}{4} = \frac{3}{2}, \quad x_2 = \frac{-16}{4} = -4
\]

### Final Solution
The correct solutions to the equation \(2x^2 + 5x = 12\) are:

\[
x = \frac{3}{2} \quad \text{and} \quad x = -4
\]