Question

\left(\frac{1}{4}x^2-3x+2.25\right)+\left(0.75x^2+2x-\frac{1}{4}\right)

Answer 1

To simplify the expression \(\left(\frac{1}{4}x^2-3x+2.25\right)+\left(0.75x^2+2x-\frac{1}{4}\right)\), we will first combine the like terms.

1. **Combine the \(x^2\) terms**:
- \(\frac{1}{4}x^2 + 0.75x^2\)
- Convert \(0.75\) to a fraction: \(0.75 = \frac{3}{4}\)
- So, \(\frac{1}{4}x^2 + \frac{3}{4}x^2 = \frac{1 + 3}{4}x^2 = \frac{4}{4}x^2 = x^2\)

2. **Combine the \(x\) terms**:
- \(-3x + 2x = -3x + 2x = -1x = -x\)

3. **Combine the constant terms**:
- \(2.25 - \frac{1}{4}\)
- Convert \(2.25\) into a fraction: \(2.25 = \frac{9}{4}\)
- So, \(\frac{9}{4} - \frac{1}{4} = \frac{9 - 1}{4} = \frac{8}{4} = 2\)

Now, putting everything together:

\[
x^2 - x + 2
\]

Thus, the simplified expression is:

\[
\boxed{x^2 - x + 2}
\]