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How do people teach factoring a quadratic equation? What are the different methods used? ---- Here's a sketch of one idea: * First, compare the results of 23*23, 22*24, 21*25, 20*26, and so on. ** See how the successive answers diverge from the originals by the squares. *** 529, 529-1, 529-4, 529-9, etc. * Try again, starting with a different number: ** 38*38, 37*39, 36*40, 35*41, 34*42, ... *** 1444, 1444-1, 1444-4, 1444-9, 1444-16, ... * Draw out the conclusion that the product of two numbers is a bit less than the square of the average ... ** and the error is the square of the distance to the average. *** 60*66 is roughly 63*63, and the error is EQN:(63-60)^2 **** So EQN:60*66=63^2-3^2=3960 * So a product, any product, /every/ product, is a square minus an error. Now let's look at EQN:x^2+12x+27. This should be a square minus the error. Let's suppose it's EQN:(x+a)^2. That's EQN:x^2+2ax+a^2. We use whatever value of *a* make the middle term work, so we want EQN:2ax=12x. So let's set a=6. That gives us EQN:(x+6)^2=x^2+12x+36, which is too big by 9. That's the square of the amount that *a* is the wrong thing to use. The square root of 9 is 3. That means we should use a+3 and a-3, which is 6+3 and 6-3, 9 and 3. * EQN:x^2+12x+27=(x+9)(x+3). We can even make this mechanical. * Example: EQN:y=x^2+6x+5 * Set *a* to be half the middle term. ** EQN:a=\frac{6}{2}=3 * Take (x+a)^2 ** EQN:(x+3)(x+3)=x^2+6x+9 * Find the error: ** EQN:E=(x+a)^2-y=(x^2+6x+9)-(x^2+6x+5)=4 * Take the square root of the error: ** EQN:\sqrt{E}=\sqrt{4}=2 * The values to use are /a-2/ and /a+2,/ which are 1 and 5. * Answer ** EQN:(x+1)(x+5) This seems a long drag, but it starts with some ~really useful number familiarisation. By playing with squares and seeing how products are more-or-less close to the square of the average, students can get a feel for numbers. We can even demonstrate by drawing a square, and seeing how the area varies if we increase one side and decrease the other. The area is nearly constant, and the error is itself a square. Diagrams to follow if someone asks. This method can even lead to the standard formula. Writing that slightly differently than usual, the solutions are /-a+e/ and /-a-e,/ where *a* is half the coefficient of *x* and *e* is the square root of the error. If the error is negative you have no real solutions. ---- Other ideas? ---- CategoryMaths