# Completing the Square Calculator

Completing the Square Calculator

The “Completing the Square Calculator” is designed to help you transform any quadratic equation of the form $ax^2 + bx + c = 0$ into a perfect square trinomial. This method is useful for solving quadratic equations and understanding the geometric properties of parabolas, such as their vertex and axis of symmetry.

### Key Features

1. Input Fields: Enter the coefficients $a$, $b$, and $c$ from the quadratic equation.

• $a$: The coefficient of $x^2$.
• $b$: The coefficient of $x$.
• $c$: The constant term.
2. Calculate Button: After entering the values of $a$, $b$, and $c$, click the “Calculate” button to perform the completing the square process.

3. Clear Button: If you want to clear the inputs and start over, click the “Clear” button.

4. Answer Section: After clicking the “Calculate” button, the solution will be displayed below, including all the steps and the final result.

### Formulas for Completing the Square

#### Step-by-Step Process

$a{x}^{2}+bx+c=0$

If $a \neq 1$, divide the entire equation by $a$ to simplify:

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
2. Move the constant $\frac{c}{a}$ to the right-hand side:

$x^2 + \frac{b}{a}x = -\frac{c}{a}$
3. Add the square of half the coefficient of $x$: Take $\left( \frac{b}{2a} \right)^2$ and add it to both sides:

$x^2 + \frac{b}{a}x + \left( \frac{b}{2a} \right)^2 = -\frac{c}{a} + \left( \frac{b}{2a} \right)^2$
4. Factor the left-hand side: The left-hand side now becomes a perfect square trinomial:

$\left( x + \frac{b}{2a} \right)^2 = -\frac{c}{a} + \left( \frac{b}{2a} \right)^2$
5. Solve for $x$: Finally, solve for $x$ by taking the square root of both sides and isolating $x$:

$x + \frac{b}{2a} = \pm \sqrt{\left( \frac{b}{2a} \right)^2 – \frac{c}{a}}$
6. The final solutions are:

$x=-\frac{b}{2a}±\sqrt{{\left(\frac{b}{2a}\right)}^{2}-\frac{c}{a}}$

### Example 1: Solving $x^2 + 2x + 1 = 0$

1. The quadratic equation is $x^2 + 2x + 1 = 0$, where $a = 1$, $b = 2$, and $c = 1$.

2. The steps are as follows:

${x}^{2}+2x=-1$

Add $\left( \frac{2}{2} \right)^2 = 1$ to both sides:

${x}^{2}+2x+1=0$

Factor the left side:

$\left(x+1{\right)}^{2}=0$

Solving for $x$, we get:

$x=-1$

Therefore, the solution is $x = -1$.

### Example 2: Solving $2x^2 + 8x – 10 = 0$

1. The quadratic equation is $2x^2 + 8x – 10 = 0$, where $a = 2$, $b = 8$, and $c = -10$.

2. First, divide the equation by 2 to simplify:

${x}^{2}+4x=5$

Add $\left( \frac{4}{2} \right)^2 = 4$ to both sides:

${x}^{2}+4x+4=9$

Factor the left side:

$\left(x+2{\right)}^{2}=9$

Solve for $x$:

$x+2=±3$

Therefore, the solutions are:

$x=1\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=-5$