Quadratic Equations
1 .- What is an equation?
2 .- What is a quadratic equation?
3 .- Solutions of a quadratic equation: Formula resolvent
4 .- Types of solutions: Real and imaginary
5 .- Examples. Verification solutions
6 .- Exercises quadratic equations are solved
1 .- What is an equation?
is an algebraic expression consisting of two separate members by an equal sign. One or both sides of the equation must have at least one variable or letter, call unknown. The equations become identities only for certain values \u200b\u200bof (s) unknown (s). These particular values \u200b\u200bare called solutions of the equation. Example:
equation: 3x - 8 = 10 only holds for X = 6, since if we substitute this value will be the identity equation: 10 = 10. So we say that X = 6 is the solution of the given equation. In fact, the only solution. If we used, for example, X = 2, be 2 = 10 (nonsense)
solve an equation is finding the values \u200b\u200bof X that satisfy through various mathematical techniques. If the equation is in first grade, a punt is the general procedure. If the degree of the equation is greater than one, should use other methods.
2 .- What is a quadratic equation?
is a particular type of equation where the unknown variable or is squared, that is, second degree. An example would be: 2x2 - 3x = 9. In this type of equation can not easily clear the X, therefore requires a general procedure to find solutions.
3 .- Solutions of a quadratic equation: Formula resolvent
The procedure is to perform changes in the algebraic equation of the quadratic equation: ax2 + bx + c = 0 until the X is clear. This procedure is not covered in this document. Solving a quadratic equation is called the resolvent formula:
The formula generates two responses: one with the + and one with the - sign before the root. Solving a quadratic equation is then limited, to identify the letters a, b and c and substitute their values \u200b\u200bin the formula solver.
It should be noted that, using the formula solver is a procedure that requires careful and requires extracting the square root of a number, either by calculator or manual process.
These difficulties make the mistake inexperienced student constantly on the solution. There are special procedures, applicable only to certain cases in which the roots can be found more easily and quickly. They are about factoring techniques.
4 .- Types of solutions: Real and imaginary
A quadratic equation can generate three types of solutions, also called roots, namely two distinct real roots
A real root (or two equal roots)
Two imaginary roots different
The approach distinguishes between these cases is the sign of the discriminant. Discriminant D is defined as:
D = b2 - 4.ac
If the discriminant is positive, then the square root is a real number and generates two distinct real roots
If the discriminant is zero, the result is zero, and both roots are the same number.
If the discriminant is negative, the square root is imaginary, producing two imaginary or complex roots. 5 .- Examples
. Verification solutions continación be resolved
A few examples will show all possible cases mentioned above. 5.1 .- Solving
- 5x2 + 13x + 6 = 0
points are identified, taking care of that equation is ordered respect to x, of degree greater than minor. With this condition we have: a = - 5, b = 13, c = 6. The following formula applies:
=
=
Since square roots are not usually stored, should be removed with a calculator, by trial or by the manual procedure. The desired root is 17, since the square of 17 is precisely 289. It is then that:
There are two different roots, one using the + sign and another sign -. Whether you call them X1 and X2 to the two solutions, which are:
Both values \u200b\u200bof x satisfy the equation, ie to replace it, produce an identity. Replacement procedure to test whether the values \u200b\u200bfound satisfy the equation is called the check is made.
Testing with x = 3. Result: -5. (3) 2 + 13. (3) + 6 = -45 + 39 + 6 = 0, as expected in the second member. Testing
X = -2 / 5, there is
Note that the fraction 20/25 was simplified to 4 / 5 before adding it to the other. As both responses produced identities, it is now certain that 2 and -2 / 5 are the roots of - 5x2 + 13x + 6 = 0
5.2 .- Solve: 6x - 2x = 9
letters can not be identified directly, since equation is messy and there is a zero on the right side of equality, therefore, make the necessary changes so that the equation is the desired shape. Transposing and changing place is: - x2 +6 x - 9 = 0. Now identify letters: a = -1, b = 6, c = -9, and applies the formula solver:
Note that the discriminant is zero, so there are two roots equal to 3, ie x1 = x2 = 3. Substituting the values \u200b\u200bin the original equation, it holds that: 6.3 - 32 = 18 - 9 = 9 thus has found the answer.
5.3 .- Solving:-6x + 13 = - x2
Again there to sort and transpose to obtain: x2-6x + 13 = 0; Identifying letters: a = 1, b = -6, c = 13. Applying the resolvent we have:
Oops! The discriminant is negative and no calculator will evaluate the square root of a negative number because this is a result which belongs to the complex numbers. Without going into details beyond the scope of this document, the root of -16 is 4i, where i is the basis of complex numbers or imagiarios, ie. The roots are then:
Separating the two answers, the solutions are: X1 = -3 + 2.i; X2 = -3 - 2.i. Require verification operations with complex numbers in rectangular form. It leaves the reader interested, investigate and verify.
6 .- Exercises quadratic equations are solved
The following exercises are approaches that generate a quadratic equation. Must first consider the logic of the problem, calling xa of the variables that the problem sets, but then must write the relations between the variables, according to approach and finally, solve the equation.
There is no general procedure for handling the logic of such problems, only experience will give the expertise to raise them. The interested reader can consult the book "Algebra" by Aurelio Baldor, considered by many as the bible of algebra.
6.1 .- The sum of two numbers is 10 and the sum of their squares is 58. Halle
both numbers are first assigned the variable x to one of the unknowns of the problem. There are two unknowns that are both numbers, as the problem does not distinguish between a and another, can be assigned either xa, eg
x = first number / / as the sum of both is 10, then necessarily the other will be:
10 - x = Second number
worth better explain this: If between your friend and you have Bs 1000, Bs and his friend is 400, how are you?, obviously, by subtracting the total minus 400, ie 1000 to 400 = B 600. If they have B x, the bill does not change, just do not know the value but in terms of x, ie you have 1000 - x
The final condition of the problem states that the sum of the squares of two numbers is 58 , then:
x2 + (10 - x) 2 = 58 This is the equation to solve
To solve it, apply some techniques of elementary algebra and then rearranged to apply the resolvent. The operation described in parentheses is the square of a binomial. It is a common misconception among students (very difficult to eradicate, by the way) to write: (a - b) 2 = a2 - b2, which is incorrect. The correct expression is: (a - b) 2 = a2 - b2 +
2.ab Developing the equation is: x2 + 102 - 2.10.x + x 2 = 58 => x2 + 100 - 20.x + x 2 = 58
Sorting and grouping: 2x2 - 20.x + 42 = 0, dividing by 2 whole equation: x2 - 10x + 21 = 0
Applying the resolvent is x1 = 3 and x2 = 7. The problem generated (apparently) two solutions, so you have to try both. Suppose we take the first (x = 3). Reviewing the initial approach, it appears that: First issue: x = 3, second number = 10 - 3 = 7.
If you take the second answer (x = 7), is: First issue: x = 7, second number = 10 - 7 = 3. In both cases, since there is no differentiation between the two numbers, the only answer is sought are numbers 3 and 7.
6.2 .- The length of a rectangular room is 3 meters longer than the width. If the width increases by 3 m and 2 m length increases, the area is doubled. Find the original floor area.
In this case, if no differentiation between long and wide, so be careful with the allocation and especially the interpretation of the variable x. This problem can easily be placed on the x either unknown, long or wide. Suppose that:
x = width of the room / / The length is 3 meters longer than the width, so:
x + 3 = length of the room. / / The area of \u200b\u200ba rectangle is the multiplication of both:
x. (X + 3) = area of \u200b\u200bthe room. Note that these are the initial data. The conditions of the problem
explain that the width increases by 3 meters and 2 meters long increases, so, after the increase are:
x + 3 = new width of the room
x + 5 = across the room again
(x + 3). (x + 5) = new living area
The new area is twice the first, so we propose the equation:
(x + 3). (x + 5) = 2. x. (X + 3)
multiplications are performed: x2 + 5x + 3x + 15 = 2x2 + 6x
It happens all the first member: x2 + 5x + 3x + 15 - 2x2 - 6x = 0
is simplified: - x2 + 2x + 15 = 0 This is the equation to solve. Solver is applied and is: x1 = 5 and x2 = - 3. The solution x = -3 is discarded, since x is the width of the room and can not be negative. Is taken as the only response that the original width was 5 meters. Looking at the initial conditions, it follows that the length is: x + 3 = 8 meters. So the original area was 8m.5m = 40 m2.
6.3 .- Find the area and perimeter of tríángulorectángulo shown. Dimensions are in meters
If the triangle is a right triangle, then it fulfills the Pythagorean Theorem: The square of the hypotenuse equals the sum of the squares of the legs. " The hypotenuse is the longest side (2x-5) and the other two are legs, it raises the equation:
(x + 3) 2 + (x - 4) 2 = (2x - 5) 2 each binomial Developing
squared, we have:
x2 + x2 + 32 + 2.3.x - 2.4.x + 42 = (2x) 2 - 2. (2x) 5 + 52 = x2 + 6x + 9 + x2 - 8x + 16 = 4x2 - 20x + 25
Regrouping: x2 + 6x + 9 + x2 - 8x + 16 - 4x2 + 20x - 25 = 0
Finally: -2 x2 + 18x = 0 This is the equation to solve
The roots of the equation are x1 = 0 and x2 = 9. The solution x = 0 is discarded, and then a leg would be -4 m, which is not possible. The solution is then x = 9. Thus, the triangle with equal sides is 12 meters and 5 meters and 13 meters hypotenuse. The area of \u200b\u200ba triangle is base times height over 2, the base and height are the two legs that are 90 °, so the area is A = 12. 5 / 2 = 30 m2. The perimeter is the sum of the sides, ie, P = 12 m + 5 m + 13 m = 30 m. Exercises
developed resolved and reviewed by: Carlos E. Utrera, Rev: June 2006
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