How to know how many solutions an equation has

All Signs Point to the Discriminant

Have you ever owned one of those Magic 8 Balls? They look like comically oversized pool balls, but have a flat window built into them, so that you can see what's insidea 20-sided die floating in disgusting opaque blue goo. Supposedly, the billiard ball has prognostic powers; all you have to do is ask it a question, give it a shake, and slowly, mystically, like a petroleum-covered seal emerging from an oil spill, the die will rise to the little window and reveal the answer to your question.

The quadratic equation contains a Magic 8 Ball of sorts. The expression b2 - 4ac from beneath the radical sign is called the discriminant, and it can actually determine for you how many solutions a given quadratic equation has, if you don't feel like actually calculating them. Considering that an unfactorable quadratic equation requires a lot of work to solve (tons of arithmetic abounds in the quadratic formula, and a whole bunch of steps are required in the completing the square method), it's often useful to gaze into the mystic beyond to make sure the equation even has any real number solutions before you spend any time actually trying to find them.

Talk the Talk

The discriminant is the expression b2 - 4ac, which is defined for any quadratic equation ax2 + bx + c = 0. Based upon the sign of the expression, you can determine how many real number solutions the quadratic equation has.

Here's how the discriminant works. Given a quadratic equation ax2 + bx + c = 0, plug the coefficients into the expression b2 - 4ac to see what results:

  • If you get a positive number, the quadratic will have two unique solutions.
  • If you get 0, the quadratic will have exactly one solution, a double root.
  • If you get a negative number, the quadratic will have no real solutions, just two imaginary ones. (In other words, solutions will contain the i you learned about in Wrestling with Radicals.)

The discriminant isn't magic. It just shows how important that radical is in the quadratic formula. If its radicand is 0, for example, then you'll get

a single solution. If, however, b2 - 4ac is negative, then you'll have a negative inside a square root sign in the quadratic formula, meaning only imaginary solutions.

Example 4: Without calculating them, determine how many real solutions the equation 3x2 - 2x = -1 has.

Solution: Set the quadratic equation equal to 0 by adding 1 to both sides.

  • 3x2 - 2x + 1= 0

You've Got Problems

Problem 4: Without calculating them, determine how many real solutions the equation 25x2 - 40x + 16 = 0 has.

Set a = 3, b = -2, and c = 1, and evaluate the discriminant.

  • b2 - 4ac
  • =(-2)2 - 4(3)(1)
  • = 4 - 12
  • = -8

Because the discriminant is negative, the quadratic equation has no real number solutions, only two imaginary ones.

How to know how many solutions an equation has

Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

You can purchase this book at Amazon.com and Barnes & Noble.

  • Algebra: The Quadratic Formula

A quadratic equation is an equation that looks like:

x2 + 4x - 2 = 0.

The general form of this is written as ax2 + bx + c = 0, where a, b and c are all numbers, and x is our unknown variable. In the example above, we would have a = 1, b = 4 and c = -2.

In order to find the number of solutions, we shall split the quadratic equation into 3 cases.

Case 1: 2 unique solutions - eg x2 + 5x + 6 = 0. Has solutions x = 2 and x = 3.

Case 2: 1 repeated solution - eg x2 + 4x + 4 = 0. Has solution x = 2.

Case 3: No solutions - eg x2 + 2x + 4 = 0. Has no solutions.

But how do we know which case we are in? To do this, we take a look at the quadratic formula, which you will hopefully have seen by now. For reference, it gives the solution of the general quadratic ax2 + bx + c = 0 as:

x = [-b ± √(b2 - 4ac)]/2a

where the ± signifies that the two solutions are 

x = [-b + √(b2 - 4ac)]/2a and 

x = [-b - √(b2 - 4ac)]/2a.

In Case 1, this will give two separate answers for x. In Case 2, both answers will be the same.

However, in Case 3 you will likely arrive at an error! This error arises from the fact that we cannot take the square root of a negative number*. This means, that if we are in case 3, then the section √(b2 - 4ac) is the part that is causing problems! As I said, we cannot take the square root of a negative number, so if b2 - 4ac is negative, we have an error, and no solutions.

This is the key to knowing how many solutions we have: 

If b2 - 4ac is positive (>0) then we have 2 solutions.

If b2 - 4ac is 0 then we have only one solution as the formula is reduced to x = [-b ± 0]/2a. So x = -b/2a, giving only one solution.

Lastly, if b2 - 4ac is less than 0 we have no solutions. 

Example:

How many solutions does x2 - 3x + 2 = -1 have?

1) Rearrange to fit the general formula: x2 - 3x + 3 = 0. So a = 1, b = -3 and c = 3.

2) Use the formula: b2 - 4ac = (-3)2 - 4(1)(3) = 9 - 12 = -3.

3) As b2 - 4ac < 0, we have no solutions.

So there you have it! Please get in touch if you require any further assistance.

For those interested/advanced students: Technically, you CAN take a square root of a negative number. It's beyond the scope of a GCSE course, so if you're confused by anything after this, don't worry! First of all though, I'll explain why nobody has told you this yet.

Imagine that I asked you to give me the answer to 7 ÷ 3, but you could only use whole numbers. The equation 7 ÷ 3 is equal to 2.33..., but this is not a whole number! So no whole number solutions exist. If I allowed you to use fractions, you could tell me that 7 ÷ 3 is 7/3 or 2 and 1/3.

The same idea applies to the problem here. We only have Real numbers (that is, fractions, decimals, whole numbers and "irrational" numbers such as pi) to deal with the question, and if you are asked to take the square root of a negative number, there are no Real solutions! 

A solution does exist in the "Imaginary" numbers. You don't know about these numbers yet (just like you didn't know about fractions at first). You will learn more about this in A level Further Maths, or perhaps at University, but if this sounds interesting please do check them out via Google.

If b2 - 4ac < 0 then there are no "Real" solutions. 

However, for your GCSEs, saying that there are no solutions will be good enough for the exam!