Question

Simplify $(\frac {1}{4}x^{2}-3x+2.25)+(0.75x^{2}+2x-\frac {1}{4})$ (1 point)
The sum of the polynomials is $\square $

Answer 1

To simplify the expression \((\frac {1}{4}x^{2}-3x+2.25)+(0.75x^{2}+2x-\frac {1}{4})\), we start by combining like terms from both polynomials.

First, let's rewrite the expression for clarity:

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

Now, let's group the like terms together:

1. **Quadratic terms** (\(x^2\)):
\[
\frac{1}{4}x^2 + 0.75x^2
\]
We convert \(0.75\) to a fraction:
\[
0.75 = \frac{3}{4}
\]
Therefore:
\[
\frac{1}{4}x^2 + \frac{3}{4}x^2 = \left(\frac{1}{4} + \frac{3}{4}\right)x^2 = \frac{4}{4}x^2 = 1x^2
\]

2. **Linear terms** (\(x\)):
\[
-3x + 2x = (-3 + 2)x = -1x = -x
\]

3. **Constant terms**:
\[
2.25 - \frac{1}{4}
\]
We convert \(2.25\) to a fraction:
\[
2.25 = \frac{9}{4}
\]
Therefore:
\[
\frac{9}{4} - \frac{1}{4} = \frac{9 - 1}{4} = \frac{8}{4} = 2
\]

Now we can piece everything together:

Combining the results from each of the groups we have:
\[
x^2 - x + 2
\]

Thus, the sum of the polynomials is:

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