Q: What formula name that gives the solutions for x in a quadratic equation?

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A quadratic equation can be solved by completing the square which gives more information about the properties of the parabola than with the quadratic equation formula.

Not necessarily. Formula gives two values but they can be identical.

Yes because rearranging it into the form of 3x2-10x+5 = 0 makes the discriminant of the quadratic equation greater than zero which means it will have two different solutions. Solving the equation by means of the quadratic formula gives x as being 2.721 or 0.613 both corrected to 3 decimal places.

2t2+8t+5 = 0 The expression in this quadratic equation is not so simple to factorise because 5 is a prime number which has only two factors (itself and one) but we can get a near enough solution by using the quadratic equation formula. Using the quadratic equation formula gives the solution as: t = - 0.7752551286 or t = - 3.224744871

Quadratic formula gives roots as 1.53 and 10.47 (to nearest hundredth.)

Completing the square is one method for solving a quadratic equation. A quadratic equation can also be solved by several methods including factoring, graphing, using the square roots or the quadratic formula. Completing the square will always work when solving quadratic equations and is a good tool to have. Solving a quadratic equation by completing the square is used in a lot of word problems.I want you to follow the related link that explains the concept of completing the square clearly and gives some examples. that video is from brightstorm.

There are no integer roots of this equation. Using the quadratic formula gives roots of 1.34 and 3.04 plus or minus loose change in each case.

If the discriminant - the part under the radical sign in the quadratic formula - is negative, then the result is complex, it is as simple as that. You can't convert a complex number to a real number. If a particular problem requires only real-number solutions, then - if the formula gives complex numbers - you can state that there is no solution.

For a quadratic equation in the form: ax² + bx + c = 0 The quadratic formula comes from completing the square of the quadratic equation that gives you a result of x=-b±√b²-4ac divided by 2a. Using a simple quadratic equation like x² + 2x + 1 = 0 a=1; b=2; c=1 x=-2±√2²-4(1)(1) divided by 2a x=-2±√4-4 divided by 2a (4 - 4 = 0 and the square root of 0 is 0) Therefore, x=-2

If you have a quadratic, which is factored like (x - P)(x - Q) = 0, so P & Q are solutions for x. Multiplying the binomials gives: x2 - Px - Qx + PQ = 0 ---> x2 - (P+Q)x + PQ = 0, so the negative of the sum is the coefficient of the x term, and the product is the constant term (no variable x).

In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.

Assuming the equation is 2b2+5b-35=0, the quadratic formula is one way to solve this equation. It can't be factored easily. Do not confuse the "b" variable in the formula given with the "b" in the related link. The values of a, b, & c are as follows to be used in the quadratic formula: a=2, b=5, c= -35. Placing these values in the quadratic formula gives: (-5+/- sqrt(52-4*2*(-35)))/(2*2). This yields variable b = (-5+/- sqrt(305))/4 or (-5+17.46425)/4 & (-5-17.46425)/4 = 3.11606 & -5.61606 which are the two roots of the given equation.

Point of intersection: (5/8, 5/2) Square both sides of the first equation and merge it with the second equation to form a quadratic equation:- 4x2-5x+25/16 = 0 Solving the above using the quadratic equation formula gives x as having two poitive roots of 5/8 and substituting this into the linear equation gives y a value of 5/2.

No answers in integers, quadratic formula gives roots as 1.31 and -0.19

The equation must have roots of x = -1 and x = 5 So: x + 1 = 0 and x - 5 = 0 Therefore: (x + 1)(x - 5) = 0 Expanding the brackets gives the equation: x2 - 4x - 5 = 0

There is no integer solution. Quadratic formula gives x = -17.66 or 5.66

No solution in integers, Quadratic formula gives roots as 1.78 and -0.28

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.

Somebody (possibly in seventh-century India) was solving a lot of quadratic equations by completing the square. At some point, he noticed that he was always doing the exact same steps in the exact same order for every equation. Taking advantage of the one of the great powers and benefits of algebra (namely, the ability to deal with abstractions, rather than having to muck about with the numbers every single time), he made a formula out of what he'd been doing:The Quadratic Formula: For ax2 + bx + c = 0, the value of x is given byThe nice thing about the Quadratic Formula is that the Quadratic Formula always works. There are some quadratics (most of them, actually) that you can't solve by factoring. But the Quadratic Formula will always spit out an answer, whether the quadratic was factorable or not.I have a lesson on the Quadratic Formula, which gives examples and shows the connection between the discriminant (the stuff inside the square root), the number and type of solutions of the quadratic equation, and the graph of the related parabola. So I'll just do one example here. If you need further instruction, study the lesson at the above hyperlink.Let's try that last problem from the previous section again, but this time we'll use the Quadratic Formula:Use the Quadratic Formula to solve x2 - 4x - 8 = 0.Looking at the coefficients, I see that a = 1, b = -4, and c = -8. I'll plug them into the Formula, and simplify. I should get the same answer as before:

No answer in integers. Quadratic formula gives roots as -1.268 and -4.732

This simplifies to x2 + 10x + 17 = 0 which has no solution in integers. Quadratic formula gives roots as -2.17 and -7.83 to the nearest hundredth. If the original equation had been "equals 0" rather than "equals 8", there would have been only one root: x = -5

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.

Doesn't have integer roots. Quadratic formula gives roots as 3.71 and -5.38.

No integer roots. Quadratic formula gives 1.55 and -0.81 to the nearest hundredth.

Are you sure you typed this correctly? Using normal methods this not factorable -- it si PRIME for the integers. You can use the quadratic formula, but that gives irrational answers for this quadratic.

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