Quadratic equation facts for kids

Kids Encyclopedia Facts

A quadratic equation is a special kind of equation that looks like this: $ax^2+bx+c=0$ . In this equation, 'a' cannot be zero. When you graph a quadratic equation, it always makes a "U" shape called a parabola on a coordinate plane.

When people work with quadratic equations, they often want to solve them. Solving means finding the points where the graph of the equation crosses the x-axis (the horizontal line). These points are also called the "zeros" or "roots" of the equation. Basically, you are trying to find the value of 'x' that makes the whole equation $ax^2+bx+c=0$ true. The numbers 'a', 'b', and 'c' are called coefficients, and they are all real numbers.

There are different ways to find the value of 'x'. One way is to graph the function. In the picture on the right, the graph crosses the x-axis at $(-1, 0)$ and $(2, 0)$ . Graphing makes it easy to see the roots, but it's not always the most convenient way. Luckily, there are other math methods to solve them!

Factoring quadratic equations
Completing the square
The quadratic formula
- How the formula is proven
- Using the quadratic formula
Images for kids
See also

Factoring quadratic equations

Some quadratic equations can be solved by factoring. Let's look at an example: $2x^2-x-1=0$ . This equation can be rewritten as $(2x+1)(x-1)=0$ .

When two things are multiplied together and the result is zero, it means at least one of them must be zero. So, we get two simpler equations: $2x+1=0$ and $x-1=0$ .

Now, it's easy to use algebra to find 'x' for each.

From $2x+1=0$ , we get Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): 2x=-1 , so $x=-1/2$ .
From $x-1=0$ , we get $x=1$ .

These two values of 'x' ( -1/2 and 1) are the roots. They make the original quadratic equation equal to zero. This means the graph crosses the x-axis at $(-1/2, 0)$ and $(1, 0)$ .

Completing the square

Not all quadratic equations can be factored easily. For example, $5x^2-4x-2=0$ is harder to factor. One way to solve these is called completing the square. This method helps us rewrite the equation so we can easily find 'x'.

Here are the steps to solve a quadratic equation by completing the square:

Step 1: Divide the entire equation by the value of 'a'. In our example, 'a' is 5.

$x^2 - \frac{4}{5}x - \frac{2}{5} = 0$
Step 2: Move the number term (the one without 'x') to the other side of the equation.

$x^2 - \frac{4}{5}x = \frac{2}{5}$
Step 3: "Complete the square" on the left side. To do this, add Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \left(\frac{b}{2a}\right)^2 to both sides. In our example, 'b' is -4/5 (from the Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x^2 - \frac{4}{5}x part, where 'a' is now 1). So, we add Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \left(\frac{-4/5}{2}\right)^2 = \left(\frac{-4}{10}\right)^2 = \left(\frac{-2}{5}\right)^2 = \frac{4}{25} to both sides.

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x^2 - \frac{4}{5}x + \frac{4}{25} = \frac{2}{5} + \frac{4}{25}
The left side can now be written as a perfect square: Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \left(x - \frac{2}{5}\right)^2 .
The right side simplifies to Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \frac{10}{25} + \frac{4}{25} = \frac{14}{25} .

So, we have: $\left(x - \frac{2}{5}\right)^2 = \frac{14}{25}$
Step 4: Find the square root of both sides. Remember to include both positive and negative roots ( $\pm$ ).

$x - \frac{2}{5} = \pm\sqrt\frac{14}{25}$
Step 5: Finally, move the number term (Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): -\frac{2}{5} ) to the other side to find 'x'.

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x = \frac{2}{5} \pm\sqrt\frac{14}{25}

This gives us two values for 'x':

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x = \frac{2}{5} + \sqrt\frac{14}{25} \approx 1.148
Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x = \frac{2}{5} - \sqrt\frac{14}{25} \approx -0.348

These are the x-coordinates where the graph crosses the x-axis. Also, the lowest or highest point of the graph, called the vertex, can be found from step 3. For this equation, the vertex is $\left(\frac{2}{5}, -\frac{14}{25}\right)$ .

The quadratic formula

While completing the square works, there's an even more general way to solve any quadratic equation. This is called the "quadratic formula." It's a special formula where you just plug in the 'a', 'b', and 'c' values from your equation.

The quadratic formula is:

x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}

Remember, 'a', 'b', and 'c' are the numbers from your original equation: Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): ax^2 + bx + c = 0 . This formula works for all quadratic equations. If 'a' were 0, it wouldn't be a quadratic equation anymore because the $x^2$ term would disappear.

How the formula is proven

The quadratic formula itself comes from using the "completing the square" method, but with 'a', 'b', and 'c' instead of actual numbers. Here's how it's done:

Step 1: Start with $ax^2+bx+c=0$ . Divide the entire equation by 'a'.

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x^2 + \frac{b}{a} x + \frac{c}{a}=0
Step 2: Move the Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \frac{c}{a} term to the other side.

$x^2 + \frac{b}{a} x= -\frac{c}{a}$
Step 3: Complete the square on the left side by adding Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \left(\frac{b}{2a}\right)^2 to both sides.

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}
The left side becomes a perfect square: Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \left(x+\frac{b}{2a}\right)^2 .
The right side can be combined into one fraction: Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \frac{b^2-4ac}{4a^2} .

So, we have: $\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$
Step 4: Take the square root of both sides. Don't forget the $\pm$ sign.

$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\ }}{2a}$
Step 5: Move the Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \frac{b}{2a} term to the other side to solve for 'x'.

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac\ }}{2a}
Combine the terms on the right side since they have the same denominator:

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}

And that's the quadratic formula!

Using the quadratic formula

Let's use the example from before: $5x^2-4x-2=0$ . Here, we can see that:

$a=5$
$b=-4$
$c=-2$

Now, plug these values into the quadratic formula: Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x=\frac{-(-4)\pm\sqrt{(-4)^2-4(5)(-2)}}{2(5)} Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x=\frac{4\pm\sqrt{16-(-40)}}{10} Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x=\frac{4\pm\sqrt{16+40}}{10} Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x=\frac{4\pm\sqrt{56}}{10}

Solving this gives us the same two values for 'x' as before:

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x \approx 1.148
Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): x \approx -0.348

The quadratic formula is very powerful because it works for every quadratic equation.

The part inside the square root, $b^2-4ac$ , is called the discriminant. It tells us about the roots of the equation:

If the discriminant is positive (greater than 0): There will be two different real values for 'x'. This means the graph crosses the x-axis at two separate points.
If the discriminant is zero (equal to 0): There is only one real value for 'x'. This means the graph just touches the x-axis at its vertex, but doesn't cross it. It's like two roots that are exactly the same.
If the discriminant is negative (less than 0): There are no real values for 'x'. On a graph, the parabola won't touch the x-axis at all. The roots in this case are complex numbers, which involve the imaginary unit 'i' (where $i = \sqrt{-1}$ ).

Images for kids

Figure 1. Plots of quadratic function y = ax2 + bx + c, showing how changing 'a', 'b', or 'c' affects the graph.
Figure 4. A graphing calculator finding one of the roots of the quadratic equation $2 x 2 + 4 x - 4 = 0$ .
A quadratic function without real roots: y = (x − 5)2 + 9. The graph does not cross the x-axis.
The path of a cliff jumper follows a parabola because of how gravity works. This path can be described by a quadratic equation.
A graph showing the difference between two ways of calculating a root for a quadratic equation.