Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Solving Quadratics with Imaginary Roots


Date: 05/28/98 at 22:53:07
From: Renier  Fernandez
Subject: Algebra

Hi, 

I go to home school and I am stuck on this problem:

   3x^2 + 2x + 5 = 0  
  
The question is, how do you solve it?

Thank you,
Renier


Date: 06/09/98 at 16:51:04
From: Doctor Peterson
Subject: Re: Algebra

Hi, Renier,

I happen to home school my own kids, so I'm happy to help.

There's a good reason for you to be stuck on this problem, because it 
doesn't have a solution - unless you are learning about imaginary 
numbers! Since I'm not sure what you're learning about, I'll give you 
several ways to look at this problem.

Your problem looks like this:

   3x^2 + 2x + 5 = 0

You may be trying to solve it by factoring. If you are, you've 
probably been trying everything you can think of and nothing works. 
That's the problem with factoring: you can tell when you've done it, 
but you can't easily tell when it can't be done - and most quadratic 
equations I could make up can't be factored. In real life we usually 
solve first, then factor if we need to by using the solution to give 
us the factors. The quickest way is to use the quadratic formula:

       -b +/- sqrt(b^2 - 4ac)
   x = ---------------------
               2a

where the equation to be solved is ax^2 + bx + c = 0.

For this problem, a = 3, b = 2, and c = 5, so the answer is:

       -2 +/- sqrt(4-60)
   x = ----------------
              6

Since 4 - 60 = -56, you can't take the square root (as a real number), 
so we know there is no solution. That saves us a lot of work 
factoring.

(If you know about imaginary numbers, you can simplify sqrt(-56) to 
4 * i * sqrt(14) and get a pair of complex solutions. If you need to 
understand that, let me know. For the rest, I'll assume we only care 
about real solutions.)

Suppose you were trying instead to solve the equation by completing 
the square. Then you would have written something like this:

   3x^2 + 2x + 5 = 0
   3(x^2 + 2/3 x) + 5 = 0               (factor 3 out of the x terms)
   3(x^2 + 2/3 x + 1/9) + 5 - 1/3 = 0   (add and subtract 1/9 to make     
                                         a square)
   3(x + 1/3)^2 + 14/3 = 0              (write it as a square)

This tells you that there is no solution, because (x + 1/3)^2 is never 
less than zero, and it would have to be -14/9 to make the left side 
zero. Do you see why this is?

What does it mean for this equation not to have a solution? Try 
graphing y = 3x^2 + 2x + 5 and you'll see that it's a parabola that 
never crosses the x axis; that is, there is no x for which it equals 
zero. In fact, if you look back at the solution by completing the 
square, the bottom of the parabola (its lowest value) is at x = -1/3, 
when (x + 1/3) is zero, and at that point y = 14/3. So even though 
completing the square didn't give you a solution, it gave us lots of 
information about the equation.

One more thought: it could be that you couldn't solve this because you 
copied the equation wrong! Make sure you check, because a change of 
sign could change everything.

I may have gone through all this too fast for you, because I'm not 
sure which parts to concentrate on. The main idea is to see how you 
can tell when a quadratic equation can't be solved. Remember: if you 
can't solve something because it can't be solved, you haven't failed! 
If you can show that it has no solution, that's the answer!

Let me know if there's more I can help with.

-Doctor Peterson,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Imaginary/Complex Numbers

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/