In this section, you can use the practice tool to improve your quadratic equation-solving skills. A quadratic equation has the form

$$ \begin{align} ax^2 + bx + c &= 0 \; \; \color{#E90F89}{\text{ or}} \\[5pt] ax^2 + bx &= c, \end{align}$$

where a, b and c are parameters.

Sometimes these equations are factorable, that is reducible to a product of binomials like $(x + d)(x + e) = 0,$ so that they're fairly easy to solve. But most of the time they are not, so you'll need to learn to solve them by either completing the square (covered here) or using the Po Shen Loh method.

The basic idea behind completing the square is that if we take a look at a perfect square of binomials, like

$$(x + m)(x + m) = x^2 + 2m(x) + m^2,$$

(m is a constant) we notice that the constant term, $m^2,$ is exactly the square of half of what multiplies $x,$ which is $2m.$ That is,

$$\left( \frac{2m}{2} \right)^2 = m^2$$

When we complete the square, we'll *force* the left side of the equation to be a perfect square, at the expense of collecting some extra constants on the right. That's OK, we can deal with those at the end.

Let's do this by example. To solve a quadratic equation like

$$3x^2 - 2x + 7 = 0,$$

which is not factorable (as far as I know – maybe you're better at it than I am), here are the steps:

__1. Move the constant to the right side.__

$$3x^2 - 2x = -7$$

__2. Divide by the coefficient of $x^2.$__

$$x^2 - \frac{2}{3} x = -\frac{7}{3}$$

__3. Add the square of ½ of the coefficient of x to both sides of the equation.__

Notice that half of $\frac{2}{3}$ is $\frac{1}{3},$ a simple thing often missed. Don't forget, as always in algebra, to add it to *both* sides.

$$x^2 \color{#E90F89}{-} \frac{2}{3} x + \left( \color{#E90F89}{\frac{1}{3}} \right)^2 = -\frac{7}{3} + \frac{1}{9}$$

__4. Identify the perfect square and clean up the right side.__

Notice, above, that the added square on the left was left un-squared. If you do that, it's very easy to ID the perfect square. It's just "x", then the highlighted sign ( - ), then the (magenta) number that's squared:

$$\left( x - \frac{1}{3} \right)^2 = \frac{-20}{9}$$

__5. Take the square root of both sides.__

$$x - \frac{1}{3} = ± \frac{2i \sqrt{5}}{3}$$

This equation has no real solutions, but we can still find the imaginary ones.

__6. Move the constant to the right and simplify the solution.__

$$x = \frac{1}{3} ± \frac{2i \sqrt{5}}{3}$$

Here we can't reduce the square root, but you should always make sure to reduce those to the smallest possible representation of the original(s).

This widget generates random quadratic equations. See if you can solve them. Then hit [**Answer**] to see the steps of the completing-the-square process. Do as many as you can. It's a terrific skill to have to get good at this. You'll likely need it as you move ahead in your math studies. **Note**: All denominators are rationalized*, so you might have to do some work to match your solution to the one shown here.

*

Hit answer to view solution.

There are a number of algebra practice pages you can use to improve your skills. Just pick the rough type of problem you need to work on and go for it.

Type 1: $ax + by = c$

Type 2: $\frac{a}{b}x + \frac{c}{d} = \frac{e}{f}$

Type 3: $\frac{a}{x} + b = c$

Type 4: $\frac{a}{x} + \frac{b}{c} = \frac{d}{e}$

Type 5: $x^2 + b = c$

Type 5: $ax^2 + \frac{b}{c} = \frac{d}{e}$

You can also learn more about quadratic functions from these pages:

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### Parameter

A parameter is an adjustable constant in the definition of a function that is different from the independent variable(s). Parameters are not independent variables. For example, in the quadratic function

f(x) = Ax^{2} + Bx + C

A, B and C are parameters which change the shape of the graph of the function. x is the independent variable. A, B and C are fixed for any particular version of f(x), but x can range from -&inf; to +&inf;

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