Quadratic equations are of the form ax2 + bx + c = 0, where a, b, and c are numeric constants and a does not equal 0 (note that a can equal 1 or -1). If you have an equation containing one variable (x) that has one or more terms of x to the second power, you can usually shift the terms in the equation around using basic algebraic operations to get 0 on one side of equals sign and ax2, etc. on the other side.  For example, let's consider the algebraic equation. 5x2 + 7x - 9 = 4x2 + x - 18 can be simplified to x2 + 6x + 9 = 0, which is in the quadratic form. Equations with greater powers of x, like x3, x4, etc. can't be quadratic equations. They are cubic equations, quadratic equations, and so on, unless the equation can be simplified to eliminate these terms of x above the power of 2. If your quadratic equation it is in the form x2 + bx + c = 0 (in other words, if the coefficient of the x2 term = 1), it's possible (but not guaranteed) that a relatively simple shortcut can be used to factor the equation. Find two numbers that both multiply to make c and add to make b. Once you find these two numbers d and e, place them in the following expression: (x+d)(x+e). These two terms, when multiplied together, produce your quadratic equation - in other words, they are your quadratic equation's factors.  For example, let's consider the quadratic equation x2 + 5x + 6 = 0. 3 and 2 multiply together to make 6 and also add up to make 5, so we can simplify this equation to (x + 3)(x + 2). Slight variations on this basic shortcut exist for slight variations in the equation itself:  If the quadratic equation is in the form x2-bx+c, your answer is in this form: (x - _)(x - _). If it is in the form x2+bx+c, your answer looks like this: (x + _)(x + _). If it is in the form x2-bx-c, you answer is in the form (x + _)(x - _).   Note: the numbers in the blanks can be fractions or decimals. For example, the equation x2 + (21/2)x + 5 = 0 factors to (x + 10)(x + 1/2). Believe it or not, for uncomplicated quadratic equations, one of the accepted means of factoring is simply to examine the problem, then just consider possible answers until you find the right one. This is also known as factoring by inspection. If the equation is in the form ax2+bx+c and a>1, your factored answer will be in the form (dx +/- _)(ex +/- _), where d and e are nonzero numerical constants that multiply to make a. Either d or e (or both) can be the number 1, though this is not necessarily so. If both are 1, you've essentially used the shortcut described above. Let's consider an example problem. 3x2 - 8x + 4 at first seems intimidating. However, once we realize that 3 only has two factors (3 and 1), it becomes easier, because we know that our answer must be in the form (3x +/- _)(x +/- _). In this case, adding a -2 to both blank spaces gives the correct answer. -2 × 3x = -6x and -2 × x = -2x. -6x and -2x add to -8x. -2 × -2 = 4, so we can see that the factored terms in parentheses multiply to become the original equation. In some cases, quadratic equations can be quickly and easily factored by using a special algebraic identity. Any quadratic equation of the form x2 + 2xh + h2 = (x + h)2. So, if, in your equation, your b value is twice the square root of your c value, your equation can be factored to (x + (sqrt(c)))2. For example, the equation x2 + 6x + 9 fits this form. 32 is 9 and 3 × 2 is 6. So, we know that the factored form of this equation is (x + 3)(x + 3), or (x + 3)2. Regardless of how you factor your quadratic expression, once it is factored, you can find possible answers for the value of x by setting each factor equal to zero and solving. Since you're looking for values of x that cause your equation to equal zero, a value of x that makes either of your factors equal zero is a possible answer for your quadratic equation. Let's return to the equation x2 + 5x + 6 = 0. This equation factored to (x + 3)(x + 2) = 0. If either of the factors equals 0, the entire equation equals 0, so our possible answers for x are the numbers that make (x + 3) and (x + 2) equal 0. These numbers are -3 and -2, respectively. When you've found your possible answers for x, plug them back in to your original equation to see if they are valid. Sometimes, the answers you find don't cause the original equation to equal zero when plugged back in. We call these answers extraneous and disregard them.  Let's plug -2 and -3 into  x2 + 5x + 6 = 0. First, -2:  (-2)2 + 5(-2) + 6 = 0 4 + -10 + 6 = 0 0 = 0. This is correct, so -2 is a valid answer.   Now, let's try -3:  (-3)2 + 5(-3) + 6 = 0 9 + -15 + 6 = 0 0 = 0. This is also correct, so -3 is also a valid answer.
تأكد من أن المعادلة في الصورة التربيعية (ax2 + bx + c = 0). في المعادلات التربيعية حيث يكون a =1، قم بتحليلها إلى (x+d )(x+e), حيث يكون حاصل ضربd × e = c و d + e = b. إذا كان الأمر ممكنًا، قم بالتحليل بالتجربة. حل عن طريق توحيد التربيع. استخدم المعاملات لحل المعادلات التربيعية. راجع إجاباتك، فبعض الإجابات قد تكون غريبة!