There are so far 8 common methods to solve quadratic equations:
A. When the equation can't be factored, the best choice would be the quadratic formula. How to know if the equation can't be factored? There are 2 ways:
1. Start solving by the new Transforming Method in composing factor pairs of a*c (or c). If you can't find the pair whose sum equals to (-b), or b, then the equation can't be factored.
2. Calculate the Discriminant D = b^2 - 4ac. If D isn't a perfect square, then the equation can't be factored.
B. When the equation can be factored, the new Transforming Method would be the best choice.
subtract
i think u can classify it in 2 ways
It is possible in only 6! = 6x5x4x3x2x1 or 720 ways.
three
5
Teachers can find many ways to teach students the quadratic equation. An activity could include having contests where students race to solve the equations in the fastest time.
First, write the equation in standard form, i.e., put zero on the right. Then, depending on the case, you may have the following options:Factor the polynomialComplete the squareUse the quadratic formula
That depends on what type of equation it is because it could be quadratic, simultaneous, linear, straight line or even differential
Four? Factoring Graphing Quadratic Equation Completing the Square There may be more, but there's at least four.
solve it
There are many ways: one is to factorise. If the quadratic is written as ax2 + bx + c then, if b2 = 4ac, the quadratic is a perfect square. It is (x - b/2a)2
The answer depends on the nature of the equation. Just as there are different ways of solving a linear equation with a real solution and a quadratic equation with real solutions, and other kinds of equations, there are different methods for solving different kinds of imaginary equations.
There are several ways to solve such equations: (1) Write the equation in the form polynomial = 0, and solve the left part (where I wrote "polynomial"). (2) Completing the square. (3) Use the quadratic formula. Method (3) is by far the most flexible, but in special cases methods (1) and (2) are faster to solve.
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You can solve a quadratic equation 4 different ways. graphing, which is quick but not reliable, factoring, completing the square and using the quadratic formula. There is a new fifth method, called Diagonal Sum Method, that can quickly and directly give the 2 roots in the form of 2 fractions, without having to factor the equation. It is fast, convenient, and is applicable whenever the equation can be factored. Finally, you can proceed solving in 2 steps any given quadratic equation in standard form. If a=1, solving the equation is much simpler. First, you always solve the equation in standard form by using the Diagonal Sum Method. If it fails to find answer, then you can positively conclude that the equation is not factorable, and consequently, the quadratic formula must be used. In the second step, solve the equation by using the quadratic formula.
Here are two ways to know if a given quadratic equations can be factored (can be solved by factoring). 1. Calculate the Discriminant D = b^2 - 4ac. When D is a perfect square (its square root is a whole number), then the given equation can be factored. 2. Solve the equation by using the new Diagonal Sum method (Amazon e-book 2010). This method directly finds the 2 real roots without having to factor the equation. Solving usually requires fewer than 3 trials. If this method fails to get the answer, then we can conclude that the equation can not be factored, and consequently the quadratic formula must be used.
You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".You solve the two equations simultaneously. There are several ways to do it; one method is to solve the first equation for "x", then replace that in the second equation. This will give you a value for "y". After solving for "y", replace that in any of the two original equations, and solve the remaining equation for "x".