Pure Quadratic Equation

A pure quadratic equation is an quadratic equation that has no linear term. So we can say that a pure quadratic equation is any equation in the form

ax2 + c = 0,      where a, c ∈ ℝ and a≠0.

Why is it so special? Because it’s really easy to solve this equation. Suppose this equation:

2x2 − 18 = 0

How should we solve it? First, we add + 18:

2x2 − 18 = 0
2x2 = 18

The quadratic term is alone now. We divide by 2:

2x2 = 18
x2 = 9

Now we ask ourself a question: what number should we squared to get nine? There are two numbers that satisfy this condition: 3 because 32 = 9 and the other number is − 3 since ( − 3)2 also equals 9. We can write down by extracting the equation:

\begin{eqnarray}x^2&=&9\\\sqrt{x^2}&=&\pm \sqrt{9}\\x&=&\pm 3\\\end{eqnarray}

Take a look in the graph of the function f(x) = 2x2 − 18. It intersects the horizontal axis exactly in points ( − 3, 0) and (0, 3).

The General Solution

We can infer the general approach how to solve any pure quadratic equation. Let’s have an equation

ax2 + c = 0

We just subtract the constant c:

ax2 + c = 0
ax2 = − c

Now we divide the equation by a:

\begin{eqnarray}ax^2&=&-c\\\\x^2&=&-\frac{c}{a}\end{eqnarray}

The final step is extracting the root:

\begin{eqnarray}x^2&=&-\frac{c}{a}\\\\\sqrt{x^2}&=&\pm \sqrt{-\frac{c}{a}}\\\\x&=&\pm \sqrt{-\frac{c}{a}}\end{eqnarray}

Now we have to distinguish three possibilities:

  • The result − (c/a) is positive. The equation has solution and it has exactly two roots. First one is − (c/a) and the second one is (c/a).
  • The result − (c/a) equals zero. The equation has a solution, there is just a single root since − (c/a) = c/a = 0
  • The result − (c/a) is negative. In this case the equation doesn’t have any solution since we cannot extract the root from negative number..

Examples

Example 1: Let’s go back to our first equation: 2x2 − 18 = 0. What are the coefficients? It’s a = 2 and c = − 18. We can use the formula

x=\pm \sqrt{-\frac{c}{a}}

to compute roots of the equation. We substitute coefficients a and c:

\begin{eqnarray}x&=&\pm \sqrt{-\frac{-18}{2}}\\\\&=&\pm \sqrt{-(-9)}\\\\&=&\pm \sqrt{9}\\\\&=&\pm 3\\\end{eqnarray}

The roots of this equation are x1 = − 3 and x2 = 3.

Example 2: Solve the equation

\frac{x^2}{3} = 0.

Ok, so first we need to identify the coefficients. We can actually rewrite the equation this way:

\frac{1}{3}\cdot x^2 = 0

Now it’s clear that the quadratic coefficients equals a = 1/3. The constant is not there thus it equals zero, c = 0. We can now use the magic formula:

\begin{eqnarray}x&=&\pm \sqrt{-\frac{0}{\frac13}}\\\\&=&\pm \sqrt{0}\\\\&=&\pm 0\\\\&=&0\\\end{eqnarray}

We hit the special case when we have just a single root, x = 0. We can draw a graph of the function f(x) = x2/3:

Graph of a function x^2/3

From the graph it’s easy to see why is there just a single root: the graph intersects with the horizontal axis in just a single point (0, 0).

Example 3: The last example:

x2 + 2 = 0.

Now we have a = 1 and c = 2. We use the magic formula again:

\begin{eqnarray}x&=&\pm \sqrt{-\frac{2}{1}}\\\\&=&\pm \sqrt{-2}\end{eqnarray}

The expression − (c/a) is negative and we cannow compute square root of a negative number. Thus this equation doesn’t have any solution. We can see it from the graph since it doesn’t intersects the horizontal axis at all: