Example 1  Solve  

Solution

Now, let’s start off here by noticing that

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

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

Now, this means that

Therefore, we can write the equation in terms of u’s instead of x’s as follows,

The new equation (the one with the u’s) is a quadratic equation and we can solve that.  In fact this equation is factorable, so the solution is,

So, we get the two solutions shown above.  These aren’t the solutions that we’re looking for.  We want values of x, not values of u.  That isn’t really a problem once we recall that we’ve defined

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

So, we have four solutions to the original equation,  and .

Example 2  Solve each of the following equations.

(a)    [Solution]

(b)    [Solution]

(c)    [Solution]

(d)    [Solution]

Solution

(a)  

Okay, in this case we can see that,

and so one of the exponents is twice the other so it looks like we’ve got an equation that is reducible to quadratic in form.  The substitution will then be,

Substituting this into the equation gives,

Now that we’ve gotten the solutions for u we can find values of x.

So, we have two solutions here  and .

[Return to Problems]

(b)  

For this part notice that,

and so we do have an equation that is reducible to quadratic form.  The substitution is,

The equation becomes,

Now, going back to y’s is going to take a little more work here, but shouldn’t be too bad.

The two solutions to this equation are  and .

[Return to Problems]

(c)  

This one is a little trickier to see that it’s quadratic in form, yet it is.  To see this recall that the exponent on the square root is one-half, then we can notice that the exponent on the first term is twice the exponent on the second term.  So, this equation is in fact reducible to quadratic in form.

Here is the substitution.

The equation then becomes,

Now go back to z’s.

The two solutions for this equation are  and  

[Return to Problems]

(d)  

Now, this part is the exception to the rule that we’ve been using to identify equations that are reducible to quadratic in form.  There is only one term with a t in it.  However, notice that we can write the equation as,

So, if we use the substitution,

the equation becomes,

and so it is reducible to quadratic in form.

Now, we can solve this using the square root property.  Doing that gives,

Now, going back to t’s gives us,

In this case we get four solutions and two of them are complex solutions.  Getting complex solutions out of these are actually more common that this set of examples might suggest.  The problem is that to get some of the complex solutions requires knowledge that we haven’t (and won’t) cover in this course.  So, they don’t show up all that often.

[Return to Problems]

Example 3  Solve .

Solution

In this case we can reduce this to quadratic in form by using the substitution,

Using this substitution the equation becomes,

This doesn’t factor and so we’ll need to use the quadratic formula on it.  From the quadratic formula the solutions are,

Now, in order to get back to x’s we are going to need decimals values for these so,

Now, using the substitution to get back to x’s gives the following,

Of course we had to use a calculator to get the final answer for these.  This is one of the reasons that you don’t tend to see too many of these done in an Algebra class.  The work and/or answers tend to be a little messy.