Solve the equation  x⁴ - 7 x² +12 = 0

Solution

Now, let's start off here by noticing that

                         x⁴  = ( x² )²

In other terms, here we can notice that the variable portion of the first term (that means ignore the coefficient) is nothing more than the variable portion of second term squared. Note as well that all we actually needed to notice here is that the exponent onto the first term was twice the exponent on the first term.

This, with the fact that third term is a constant, refer that this equation is reducible to quadratic in form.  We will solve it by first defining,

                                                           u = x²

Now, It means that

                                    u ²  = ( x² )²  = x⁴

 Thus, we can write down the equation in terms of u's rather then x's as follows,

                     x⁴ - 7 x² + 12 = 0

⇒         u ² - 7u + 12 =0

The new equation (the one along with the u's) is a quadratic equation & we can solve that.  In fact this equation is factorable, thus the solution is,

u ² - 7u + 12 = (u - 4) (u - 3) = 0

⇒         u= 3, u = 4

Thus, we get the two solutions illustrated above.  These aren't the solutions which we're looking for.  We desire values of x, not values of u. That isn't actually a problem once we remember that we've defined

                                                     u = x²

To get values of x for the solution all we have to do is plug in u in this equation and solve out that for x.   Let's do that.

u = 3:   3 = x²   ⇒ x = ±√3

u = 4:   4 = x²   ⇒x=±√4 = ±2

Thus, we have four solutions to the original equation,

x = ±2 and              x = ±  √3 .

Thus, the basic procedure is to check that the equation is reducible into quadratic in form then make a rapid substitution to turn it into a quadratic equation. We solve out the new equation for u, the variable from the substitution, & then employ these solutions and the substitution definition to obtain the solutions to the equation that we actually want.

In most of the cases to make the check which it's reducible to quadratic in form all that we actually have to do is to check that one of the exponents is twice the other. There is one exception that we'll see here once we get into a set of examples.

Also, once you get "good" at these you often don't really need to do the substitution either.  We will do them to make sure that the work is clear.  However, these problems can be done without the substitution in many cases.