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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
What are the steps to solving a quadratic equation?
In general, there are two steps in solving a given quadratic equation in standard form ax^2 + bx + c = 0. If a = 1, the process is much simpler. The first step is making sure that the equation can be factored? How? In general, it is hard to know in advance if a quadratic equation is factorable. I suggest that you use first the new Diagonal Sum Method to solve the equation. It is fast and convenient and can directly give the 2 roots in the form of 2 fractions. without having to factor the equation. If this method fails, then you can conclude that the equation is not factorable, and consequently, the quadratic formula must be used. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009) The second step is solving the equation by the quadratic formula. This book also introduces a new improved quadratic formula, that is easier to remember by relating the formula to the x-intercepts with the parabola graph of the quadratic function.
What are some hard math equations that equals 7?
How about finding the solutions of the quadratic equation: x^2-14x+49 = 0
How do you solve for x in the equation x2-2x-2 equals 0?
To solve for x in the equation x2 - 2x - 2 = 0, use the quadratic equation, that is: For 0 = ax2 + bx + c, the roots (values of x) are defined as: x = [-b +/- sqrt(b2 - 4ac) ] / 2a (If that's hard to understand, google "quadratic formula"). It works out to x = 2.73, -0.73.
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 are two algebraic methods for solving quadratic equations?
Finally, there are two methods to use, depending on if the given quadratic equation can be factored or not. 1.- The first one is the new Diagonal Sum Method, recently presented in book titled: "New methods for solving quadratic equations" (Trafford 2009). This method directly gives the two roots in the form of two fractions, without having to factor it. The innovative concept of this new method is finding 2 fractions knowing their product (c/a) and their sum (-b/a). This new method is applicable to any quadratic equation that can be factored. It can replace the existing trial-and-error factoring method since this last one contains too many more permutations. In general, it is hard to tell in advance if a given quadratic equation can be factored. However, if the new method fails to get the answers, then you can positively conclude that this equation can not be factored. Consequently, the quadratic formula must be used in solving. We advise students to always try to solve the given equation by the new method first. If the student gets conversant with this method, it usually take less than 2 trials to get answers. 2. the second one uses the quadratic formula that students can find in any algebra book. This formula must be used for all quadratic equations that can not be factored.
What is an example of chain migration in history?
This means push really hard up in there ;)
What multiplication equation equals 900?
You might think it looks hard but it's really not all you have to do is say 90 x 10 = 900
What is a simple equation for the trajectory of catapults?
you find the hard equation and simplify it....
What is the standard form for the equation (-6x1y) m-4?
Well, it is hard to answer your question . . . "(-6x1y)m-4" is not an equation, you see. To be an equation, the formula needs to have an equals sign. Example: 6x+27 -17 = 32 is an equation and can be solved. However, your stated formula can be better written like this: m-4 (-6y)
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 hard is iron?
its like really really hard
What do you use the quadratic formula for?
One would use the quadratic formula for solving binomials that are otherwise hard to factor. You can find both real and imaginary solutions using this method, making it highly superior to factoring in this regard.
