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/
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