To rewrite $2x^2 + x - 3$ in the form $a(x + b)^2 + c$, we need to complete the square. Here's the step-by-step process:

Step 1: Factor out the coefficient of $x^2$ from the first two terms

We start with the given quadratic expression: $2x^2 + x - 3$ Factor out the 2 from the first two terms: $2(x^2 + \frac{1}{2}x) - 3$

Step 2: Complete the square inside the parentheses

Now, we complete the square for the expression inside the parentheses, $x^2 + \frac{1}{2}x$. To do this, take half of the coefficient of $x$, square it, and add it and subtract it inside the parentheses.

• Half of $\frac{1}{2}$ is $\frac{1}{4}$
• Squaring $\frac{1}{4}$ gives $\frac{1}{16}$

So, we add and subtract $\frac{1}{16}$ inside the parentheses: $2 \left( x^2 + \frac{1}{2}x + \frac{1}{16} - \frac{1}{16} \right) - 3$ Now, simplify: $2 \left( \left(x + \frac{1}{4}\right)^2 - \frac{1}{16} \right) - 3$

Step 3: Distribute the 2

Distribute the 2 to both terms inside the parentheses: $2 \left( x + \frac{1}{4} \right)^2 - 2 \times \frac{1}{16} - 3$ $2 \left( x + \frac{1}{4} \right)^2 - \frac{1}{8} - 3$

Step 4: Combine constants

Now, combine $-\frac{1}{8}$ and $-3$. To do this, convert $-3$ to a fraction with a denominator of 8: $-3 = -\frac{24}{8}$ Now, combine: $-\frac{1}{8} - \frac{24}{8} = -\frac{25}{8}$

Final Answer

Thus, the expression $2x^2 + x - 3$ in the form $a(x + b)^2 + c$ is: $2 \left( x + \frac{1}{4} \right)^2 - \frac{25}{8}$

Would you like a more detailed explanation or to go over any of the steps?

Related Questions:

1. What is the general method for completing the square?
2. How do you complete the square for expressions that don't have a leading coefficient of 1?
3. How do you solve quadratic equations using the method of completing the square?
4. How can you derive the quadratic formula from completing the square?
5. How does completing the square relate to graphing parabolas?

Tip: When completing the square, remember that the value you add inside the parentheses must be balanced by subtracting the same value outside to keep the equation equivalent.