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 .
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(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 .
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(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 onehalf, 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
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(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]
