Factoring expressions and Solving Equations That Are Quadratic in Form
You have already learned how to factor quadratic expressions in the form of ax2 + bx + c where a
โ 0. However, there are instances where you will be faced with trinomials of a degree higher
than two but still fit the quadratic form. For example, the trinomial ax4 + bx2 + c has a degree
power of four but it still fits the quadratic form.
ax4 + bx2 + c = a(x2)2 + b(x2) + c
= au2 + bu + c
This substitution pattern can also be applied to situations where there is an algebraic expression
in place of x in the quadratic form. Take for example the expression 2(x โ 3)2 โ 5(x โ 3) โ 12. In
this trinomial instead of have x and x2 we have (x โ 3) and (x โ 3)2.
2(x โ 3)2 โ 5(x โ 3) โ 12 = 2(x โ 3)2 โ 5(x โ 3) โ 12
= 2u2 โ 5u โ 12
The following steps can be used to solve equations that are quadratic in form:
1. Let u equal a function of the original variable (normally the middle term)
2. Substitute u into the original equation so that it is in the form au2 + bu + c = 0
3. Factor the quadratic equation using the methods learned earlier
4. Solve the equation for u
5. Replace u with the expression of the original variable
6. Solve the resulting equation for the original variable
7. Check for any extraneous solutions
Example 1: Solve the equation x4 โ 13x2 + 36 = 0.
Solution
Step 1: Let u equal a function of the original variable
In this problem, we would let u equal x2
Step 2: Substitute u into the original equation for the variable expression
Before performing the substitution rewrite x4 as a multiply of x2 which will be
replaced by u. x4 = (x2)2
x
4 โ 13x2 + 36 = 0
(x
2)2 โ 13x2 + 36 = 0
u
2 โ 13u + 36 = 0
Math 0303
Student Learning Assistance Center - San Antonio College
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