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How many existing methods are there in solving quadratic equations?
There are 5 existing methods in solving quadratic equations. For the first 4 methods (quadratic formula, factoring, graphing, completing the square) you can easily find them in algebra books. I would like to explain here the new one, the Diagonal Sum Method, recently presented in book titled:"New methods for solving quadratic equations and inequalities" (Trafford 2009). It directly gives the 2 roots in the form of 2 fractions, without having to factor the equation. The innovative concept of the method is finding 2 fractions knowing their Sum (-b/a) and their Product (c/a). It is very fast, convenient and is applicable whenever the given quadratic equation is factorable. In general, it is hard to tell in advance if a given quadratic equation can be factored. However, if this new method fails to find the answer, then we can conclude that the equation can not be factored, and consequently, the quadratic formula must be used. This new method can replace the trial-and-error factoring method since it is faster, more convenient, with fewer permutations and fewer trials.
What is a simple equation for the trajectory of catapults?
you find the hard equation and simplify it....
Is honors geometry hard?
I don't think it is but it really depends on the student. For example, I have a friend who loved the class and thought it was really easy, and I have a friend who thought it was hard and was glad to be done with it. If you work hard and don't give up or go the easy way out, then you shouldn't have a problem with it.
How do you solve the quadratic equation x2 minus 6x plus 6 equals 0?
The reason you use the Quadratic method is because certain equations can't be factored. So let's start by defining the Quadratic Formula.x = [-b +- √b2 - 4ac]/2aFor this equation, there are no factors of 6 that when added will equal 6, so we need to use the Quadratic Formula. Now let's find a, b, and c. The equation x2 - 6x + 6 = 0 is in the Standard Form of ax2 + bx + c = 0, so we just need to compare.a = 1, b = -6, and c = 6Now that we have a, b, and c defined, fill in the Quadratic Formulax = [-(-6) +- √(-6)2 - 4(1)(6)]/2(1)x = [ 6 +- √36 - 24]/2x = [ 6 +- √12]/2x = [ 6 +- 2√3]/2x = (6/2) +- [(2√3)/2]x = 3 +- √3So, you have two possible answers: 3 + √3 and 3 - √3. Now you need to check each answer to make sure both are valid.WARNING: It's hard to type complex fractions in Wiki Answers, so any clarification needed, let me know.
what is the most hard maths equation?
when u are in the high level
