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.

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.

Let its sides be x and rearrange the diagonal formula into a quadratic equation:- So: 0.5(x^2-3x) = 252 Then: x^2-3n-504 = 0 Solving the quadratic equation: gives x a positive value of 24 Therefore the polygon has 24 sides irrespective of it being irregular or regular

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.

2x²-x=0.5 is more manageable in the standard quadratic form : 2x2 - x - 0.5 = 0double this to get rid of that pesky 0.5 : 4x2 - 2x -1 = 0The quadratic formula solves ax2 + bx + c =0 using x = (-b +/- (b2-4ac) 0.5 ) / 2ax equals [minus b plus or minus (square root of {b squared minus 4ac})] all over 2aputting in a=4, b=-2, c=-1 gives (2 +/- (4 + 16)0.5 ) / 8which reduces to( 1 +/- (5)0.5) / 4this evaluates to x= .809 or -.309Notice that plus or minus square root (b2 - 4ac) usually produces two different solutions. Equations in the second degree always have two solutions; if the quadratic is such that b2 equals 4ac the formula seems to give only one solution. Don't worry about this; there are still two solutions, they just happen to be identical!

Related questions

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.

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.

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.

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

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 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).

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.

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