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### Solving Quadratics with Imaginary Roots

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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
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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.

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

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

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/
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Associated Topics:
High School Basic Algebra
High School Imaginary/Complex Numbers

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