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Why do you set each factor equal to zero when solving a quadratic equation by factoring?
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Can a factor pair have an extra addition problem?
Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.
Can Quadratic equations that can be solved using the quadratic formula also be solved by factoring?
Yes, however not all quadratic equations can easily be solved by factoring, sometimes you can factor and sometimes it is easier to use the quadratic formula. Example: x2 + 4x + 4 This can be easily factored to (x + 2)(x +2) Therefore the answer is -2 by setting x +2 = 0 and solving for x This can be done using the quadratic equation and you would get the same results, however, it was much faster to factor instead.
Can all quadratic equations be solved?
Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.
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.
How do you solve S squared plus 4s minus 21 equals to 0?
To solve the quadratic equation, S^2 + 4S - 21 = 0, you can factor the expression or use the quadratic formula. Factoring, we can rewrite it as (S-3)(S+7) = 0. This means that either S-3 = 0 or S+7 = 0. Solving for S in each case gives S = 3 or S = -7 as the solutions to the equation.
Can a factor pair have an extra addition problem?
Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.
Can Quadratic equations that can be solved using the quadratic formula also be solved by factoring?
Yes, however not all quadratic equations can easily be solved by factoring, sometimes you can factor and sometimes it is easier to use the quadratic formula. Example: x2 + 4x + 4 This can be easily factored to (x + 2)(x +2) Therefore the answer is -2 by setting x +2 = 0 and solving for x This can be done using the quadratic equation and you would get the same results, however, it was much faster to factor instead.
Can you solve a quadratic equation without factoring?
using the quadratic formula or the graphics calculator. Yes, you can do it another way, by using a new method, called Diagonal Sum Method, that can quickly and directly give the 2 roots, without having to factor the equation. This method is fast, convenient and is applicable to any quadratic equation in standard form ax^2 +bx + c = 0, whenever it can be factored. It requires fewer permutations than the factoring method does, especially when the constants a, b, and c are large numbers. If this method fails to get answer, then consequently, the quadratic formula must be used to solve the given equation. It is a trial-and-error method, same as the factoring method, that usually takes fewer than 3 trials to solve any quadratic equation. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009)
Which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring?
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.
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.
Can all quadratic equations be solved?
Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.
What is the quadratic equation used for?
It is used to solve quadratic equations that cannot be factored. Usually you would factor a quadratic equation, identify the critical values and solve, but when you cannot factor you utilize the quadratic equation.
How do you factor 5a2-14a-30?
solving by factoring 5a2
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.
How do you do a parabola?
A parabola is a graph of a 2nd degree polynomial function. Two graph a parabola, you must factor the polynomial equation and solve for the roots and the vertex. If factoring doesn't work, use the quadratic equation.
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.
How do you solve S squared plus 4s minus 21 equals to 0?
To solve the quadratic equation, S^2 + 4S - 21 = 0, you can factor the expression or use the quadratic formula. Factoring, we can rewrite it as (S-3)(S+7) = 0. This means that either S-3 = 0 or S+7 = 0. Solving for S in each case gives S = 3 or S = -7 as the solutions to the equation.
