Incomplete quadratic equations are a special case of second-order equalities. It is necessary to be able to solve these equations, since they are often found not only in mathematical, but also in physical problems. This article is devoted to methods for solving them.
Quadratic equations: complete and incomplete
Before disassembling methods for solving incomplete quadratic equations, you should consider what they are.
The figure below shows the general form of second-order equalities, which are so called because of the maximum value of the degree of the variable (it is equal to 2) contained in them.
Where a, b and c are numbers (coefficients). An incomplete equation is obtained when one of these coefficients becomes equal to zero (except for the number a, because if it vanishes, the equation will cease to be square). Since there are only three possible combinations of zero coefficients, the following types of incomplete second-order equalities are distinguished:
- Only b = 0. Then the equation is transformed to the form a * x 2 + c = 0. It is called a pure or simple incomplete equality of a square type.
- Only c = 0. Then we get the form: a * x 2 + b * x = 0. It is called the mixed incomplete quadratic equation.
- Finally, if b = 0 and c = 0, then we have the expression a * x 2 = 0.
The last form of an incomplete equation is not considered in any mathematical course, since its solution is obvious and only possible: x = 0.
Is it possible to solve incomplete equations using the discriminant formula?
Yes, you can, because this method is universal for any second-order expression. However, incomplete quadratic equations in the 8th grade of the school are already found, and they begin to be studied earlier than complete equalities of this type, for which a formula with discriminant is already given. In addition, the type of equalities under consideration is simple enough to apply universal formulas to them and make a series of unnecessary calculations.
Consider simple and understandable methods for solving incomplete second-order equations.
Solving a simple incomplete equation
The scheme for solving it is generally presented in the figure below.
Let us explain in more detail each step marked on it. The first step is to bring the equation to the form indicated at the beginning of this scheme. The condition of the problem can be so composed that the original equality will contain more than two terms. All of them must be simplified (multiply, add and subtract) to the form of pure incomplete equality.
After that, the free term c is transferred to the right side of the equality and is divided by the coefficient a. To get the unknown x, it remains to take the square root of the -c / a relationship, but one must not forget and take into account that it can be either with a minus sign or a positive sign.
What follows from the formula shown in the figure? Firstly, the roots of pure incomplete square equality are always 2-a, while they are equal in absolute value and differ in sign. Secondly, if the numbers c and a have the same sign, then the roots of x will be imaginary, if c and a are of different signs, then two real solutions are obtained.
Solving a Mixed Incomplete Equation
To solve a quadratic equation with c = 0, the same first step should be taken as in the case of determining the roots of a pure incomplete equality, that is, bring it to a form with two terms: one of them should contain x 2 , and the other x. Then, apply the factorization method, that is, factor the left side of the equality. In contrast to the complete equation, this is very simple, since one of the factors will always be x. The above can be written as a formula:
x * (a * x + b) = 0.
This equality has a solution if each of its factors is zero. The result of calculating the roots is presented in the figure below.
Thus, the roots of this type of incomplete equation will always be real numbers, and one of them is equal to zero. The sign of the second root is determined by the ratio of nonzero coefficients b / a.
Examples of math problems
Now we give illustrative examples of square incomplete equations with a solution.
Example 1. Find the roots of the equality 135- (2x + 3) (2x - 3) = 0. Expand the brackets, we get: 135-4 * x 2 + 9 = 0. Note that terms containing x in the first degree are reduced. Performing the transfer of free terms to the right side and dividing them by -4, we get: x 2 = 36. From where two roots follow: 6 and -6.
Example 2.23 * (x 2 -2) = 34 * x-46. As in the first case, we open the brackets and transfer all the terms to the left. We have: 23 * x 2 -46-34 * x + 46 = 0. Now we reduce the free terms and factor the sum, we get: x * (23 * x-34) = 0. Whence it follows that x = 0 and x = 34 / 23≈1.47826.
The solution of examples showed that the algorithm for finding the roots of any kind of an incomplete second-order equation is quite simple, so there is no point in remembering the formulas presented in the figures above.
Physical task example
Many students heard from their physics teacher that Galileo Galilei in the 17th century conducted experiments to calculate the acceleration of gravity by dropping various bodies from a tower in Pisa. This will seem curious to many, but there is no historical evidence that the scientist really conducted such experiments. However, in the same XVII century they were executed by another Italian.
Giovanni Riccioli is an astronomer and a Jesuit who was able to really calculate the acceleration of a free fall by dropping clay balls from the height of the Asinelli tower, located in the city of Bologna. Riccioli received an acceleration value of 9.6 m / s 2 (the current value is 9.81 m / s 2 ). Knowing this number, it is necessary to determine how long the clay ball fell to the ground, given that the height of the tower is 97.6 meters.
To solve the problem, it is necessary to remember that the path with uniformly accelerated motion is expressed by the formula: l = v 0 * t + g * t 2/2. Since at the moment when Riccioli released the ball, the speed of the latter was zero, then the term v 0 * t = 0. Then we come to the equation: 97.6 = 9.6 * t 2/2. Whence we get that t = 4.51 seconds (the negative root was deliberately discarded).