The Discriminant of Quadratic EquationsDate: 08/06/98 at 10:18:07 From: Quincy Subject: Quadratic equation Will you explain why b^2 - 4ac = a triangle? Date: 08/06/98 at 11:41:58 From: Doctor Santu Subject: Re: Quadratic equation Hello, Quincy! This is all about how to solve equations such as: 40 x^2 + 5 x + 63 = 0 right? People learned how to solve lots and lots of equations like this and then, a few hundred years ago, somebody asked what do the 40, the 5, and the 63 have to do with it? Can we find rules about what the answer is for every possible combination of the three "coefficients"? [In our example, 40 is the x^2-coefficient, 5 is the x-coefficient, and 63 is the constant term.] So one day, they set out to solve the equation: a x^2 + b x + c = 0. But the answers they got depended (obviously) on a, b and c, the coefficients. There were several different ways the problem could turn out, and, most interestingly, exactly which way it went was decided by this famous b^2 - 4ac. So they wanted to call that formula by a name, and the name that was decided was "discriminant," because it discriminated between the various solutions. But, as you know, in math we like to call things by single letters, like a, b, f, x and so on. So we use the Greek letter Delta as its symbol. Delta looks exactly like a triangle, but it is a Greek letter, not a geometric shape (just as the letter o looks like a circle to people who don't know what it is). The expression ax^2 + bx + c is an interesting one. With some careful math, it can be split into two parts: a * [varying part that's never negative] - [fixed part] The fixed part is Delta/[4a], so you see the connection! If Delta is negative, the whole combination has absolutely no chance of being zero, so the equation ax^2 + bx + c = 0 cannot be solved. To see this, put the whole thing over a common denominator, and you get: 4a^2 * [varying part, never negative] - Delta ----------------------------------------------- 4a If Delta is negative, -Delta will be positive, and 4a^2 is certainly positive, so the numerator is going to be at least as big as -Delta, so it can never be zero, regardless of the denominator. If Delta is positive, there will be two different values for x at which the whole expression becomes zero. This is the so-called generic case. To see this, I have to tell you that the "varying part" is just (x + b/(2a))^2, which takes any desired positive value exactly twice. If Delta is zero, the only value of x that will make the expression zero is x = -b/(2a), so you have one solution. A much nicer way of understanding the solution is to think of it in terms of graphs. You must know what the graph of the expression: ax^2 + bx + c looks like, in which case the three cases become obvious. Look in your textbook. There should be diagrams showing you the situation. There will be a u-shaped curve, called a parabola. There will be a horizontal line that's the x-axis. Each point on the x-axis stands for a possible x-value. The corresponding value of ax^2 + bx + c is just how high the point on the parabola is from the x-axis. When the curve actually crosses the x-axis, the height is zero, of course, so the value is zero. The question then becomes does this graph cross the x-axis? At how many points? The a, b, and c determine the position of the parabola. The "a" determines whether the parabola is arch-shaped (a is negative) or u-shaped (a is positive). Delta determines whether the parabola crosses the x-axis (Delta is positive), just touches the x-axis (Delta is zero) or misses the x-axis completely (Delta is negative). Thanks for writing, and write back if you want more information - Doctor Santu, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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