How to solve an incomplete quadratic equation? It is known that it is a particular variant of the equality ax 2 + bx + c = o, where a, b and c are real coefficients for unknown x, and where a ≠ o, a b and c are zeros - simultaneously or separately. For example, c = o, ≠ o or vice versa. We almost remembered the definition of a quadratic equation.
Specify
The trinomial of the second degree is zero. Its first coefficient a ≠ o, b and c can take any values. The value of the variable x will then be the root of the equation when , when substituted, it turns it into the correct numerical equality. Let us dwell on real roots, although complex numbers can also be solutions to the equation . Complete is called an equation in which none of the coefficients is equal to o, but ≠ o, in ≠ o, with ≠ o.
Let's solve an example. 2x 2 -9x-5 = o, we find
D = 81 + 40 = 121,
D is positive, so there are roots, x 1 = (9 + √121): 4 = 5, and the second x 2 = (9-√121): 4 = -o, 5. Verification will help ensure that they are correct.
Here is a phased solution to the quadratic equation
Through the discriminant, one can solve any equation on the left side of which the known square trinomial for a . In our example. 2x 2 -9x-5 = 0 (ax 2 + in + c = o)
- First we find the discriminant D by the well-known formula in 2 -4ac.
- We check what the value of D will be: we have more than zero, it can be zero or less.
- We know that if D ›, the quadratic equation has only 2 different real roots, they are denoted by x 1 usually and x 2 ,
this is how they calculated:
x 1 = (-b + √D) :( 2a), and the second: x 2 = (-b-√D) :( 2a). - D = o - one root, or, they say, two equal:
x 1 is equal to x 2 and equal to -b: (2a). - Finally, D ‹o means that the equation has no real roots.
Consider what are the incomplete equations of the second degree
- ax 2 + in = o. The free term, coefficient c at x 0 , is equal to zero, in ≠ o.
How to solve an incomplete quadratic equation of this kind? Put x out of the brackets. We recall when the product of two factors is zero.
x (ax + b) = o, it can be when x = o or when ax + b = o.
Having solved the 2nd linear equation, we have x = -b / a.
As a result, we have the roots x 1 = 0, by calculations x 2 = -b / a . - Now the coefficient of x is equal to o, and c is not equal to (≠) o.
x 2 + c = o. Move c to the right side of the equality, we get x 2 = -. This equation only has real roots when -c is a positive number (c ‹o),
x 1 then equals √ (-s), respectively x 2 - -√ (-s). Otherwise, the equation has no roots at all. - The last option: b = c = o, that is, ax 2 = o. Naturally, such a simple equation has one root, x = o.
Special cases
We considered how to solve an incomplete quadratic equation, and now we can take any kind.
- In the full quadratic equation, the second coefficient at x is an even number.
Let k = o, 5b. We have formulas for calculating the discriminant and roots.
D / 4 = k 2 - ac, the roots are calculated as x 1,2 = (-k ± √ (D / 4)) / for D ›o.
x = -k / a at D = o.
There are no roots at D ‹o. - There are given quadratic equations when the coefficient of x squared is 1, it is customary to write x 2 + px + q = o. All of the above formulas apply to them, the calculations are somewhat simpler.
Example, x 2 -4x-9 = 0. Calculate D: 2 2 +9, D = 13.
x 1 = 2 + √13, x 2 = 2-√13. - In addition, the Vieta theorem is easily applied to the above ones . It says that the sum of the roots of the equation is –p, the second coefficient with a minus (meaning the opposite sign), and the product of the same roots will be q, the free term. Check how easy it would be to verbally identify the roots of this equation. For the unreduced (with all coefficients not equal to zero) this theorem is applicable as follows: the sum x 1 + x 2 is equal to -b / a, the product x 1 · x 2 is equal to s / a.
The sum of the free term c and the first coefficient a is equal to the coefficient b. In this situation, the equation has at least one root (easy to prove), the first is necessarily -1, and the second is c / a, if it exists. How to solve an incomplete quadratic equation, you can check yourself. As easy as pie. Coefficients may be in some ratios among themselves.
- x 2 + x = o, 7x 2 -7 = o.
- The sum of all coefficients is equal to o.
The roots of this equation are 1 and s / a. Example, 2x 2 -15x + 13 = o.
x 1 = 1, x 2 = 13/2.
There are a number of other methods for solving various equations of the second degree. Here, for example, is a method for extracting a full square from a given polynomial. There are several graphic ways. When you often deal with such examples, you will learn to “click” them like seeds, because all methods come to mind automatically.