6.4 Equations Quadratic in Form

Learning Outcomes

  • Solve equations reducible to quadratic form including using uu-substitution.

In this section we will learn to factor expressions which may not appear factorable at first, but after making a substitution they become factorable using our standard techniques.

As a warm-up, consider the following expression to factor:

x425x425

Both terms are squares since x4=(x2)2x4=(x2)2, so we may use the difference of squares formula to factor this. However, we will first make a substitution to make it more clear that the formula applies. The substitution suggested by our above observation is u=x2u=x2.  After making this substitution, our expression becomes

x425=(x2)225=u225=(u+5)(u5)=(x2+5)(x25)x425=(x2)225=u225=(u+5)(u5)=(x2+5)(x25)

Make sure to write your final answer using the same variable that the problem started with.

Let’s use this technique now to solve an equation.

Example

Solve x43x24=0x43x24=0.

Since the equation was possible to change into quadratic by an appropriate substitution, we say that the original equation was quadratic in form.

In our remaining examples we will show more examples of substitutions that result in a quadratic equation to solve. Observe that it can be helpful to write the starting expression in descending order (in some sense of the word “descending”) and using the middle term to determine the correct substitution.

Example

Solve x2+5x1+6=0x2+5x1+6=0.

Note that sometimes there are alternative approaches to solving these problems. We could have written our starting problem as 1x2+5x+6=0 and solved using our rational equation methods of the last chapter, multiplying both sides by the LCM of x2 to obtain 1+5x+6x2=0. You can verify that this results in the same solutions found above.

Example

Solve xx=2.

Summary

In this section, we used a substitution to turn a variety of different equations into quadratic equations which could be solved either by factoring or using the previous methods of this chapter.