Solving Equations of Quadratic Form
An equation is said to be in quadratic form if it can be written where z is any algebraic expression. When z is simply a single variable, say x, then the equation is merely a quadratic equation and solving such an equation is the topic of several other tutorials. By broadening our choice of expressions for z we may complicate the equation considerably. However, as long as we are dealing with an equation of quadratic form, we can still apply all the techniques we have for solving quadratic equations and often simplify the matter immensely. Here are some examples of equations of quadratic form:
12m4 – 11m2 + 2 = 0 This is of the form where z = m2.
6p-2 + p-1 – 2 = 0 This is of the form where z = p-1.
This is of the form where z = (r -1)1/3
This is of the form where z = .
Solve the equation:
While there is a formula for solving polynomial equations of degree four (just as the quadratic formula can be applied to any quadratic equation), it is quite cumbersome. In this case, though, we are fortunate when it comes to the form of this equation. With a little thought, it is not too hard to see that if we were to define a new variable z, such that , the equation reduces to the following:
Now we just treat this as we would any quadratic equation and solve for z. We can always use the quadratic equation, but in this case the left hand side factors easily into:
Solving , yields the solutions and .
But, we’re not done yet! Remember, we want to solve the original equation for the variable x. Generating such solutions is the same as substituting into our new equations and since these are true if and only if the original equation is true.
When those substitutions are made, we get:
Solving both of these equations give us the following four solutions:
Note that the symbol is a shorthand way to account for both the positive and negative square roots. You can check that all four of these are solutions to the original equation by plugging them in and confirming that the right hand side equals 0 in each case.
Here’s another more difficult example that may require you to review some of the tutorials dealing with the exponential and logarithmic functions.
Solve the equation:
Let . Then the equation can be rewritten in terms of z as:
Now as in the last example, we factor the right hand side and get:
The solutions to are and . After substituting into these two equations, we get:
The solution to the first exponential equation is and the solution to the second is . You may check that these are solutions to the original equation by substitution. Remember that this method only works when dealing with an equation of quadratic form. For this reason, we could not use this method on an equation such as .