Solving Quadratic (2nd degree) Equations
There are 3 widely used methods for solving quadratic equations. A quadratic (or second-degree) equation is an equation in which the variable has an exponent of 2.
The standard form of a quadratic equation is
.
The standard form of a quadratic equation is
.
The three methods used to solve quadratic equations are: 1) factoring, 2) the square root property, and 3) the quadratic formula. Quadratic equations generally have 2 solutions.
1) Factoring is one method used to solve second-degree and larger equations. First the equation must be written in standard form. This means that the polynomial must be in descending form and set equal to zero. Next, you must factor the polynomial. (You may want to review factoring.) Once the polynomial is factored, set any factor which contains a variable equal to zero and solve (using isolation) for the variable. Check your answer in the original equation. Every quadratic equation has two solutions, although in the case of a perfect square trinomial both of the solutions are the same.
If the equation is factored and set equal to zero as in the example, , then and .
Solve the equation, . First, you must get the equation in standard form.
This means that you must add 16 to both sides so that the –16 is removed from the left side of the equation and the equation will then be equal to zero … … . Now you are ready to factor, . Set the factors equal to zero. Since is repeated twice as a factor, there are two solutions, but they are both the same. Thus, is the only “unique” solution to this problem. This is a perfect square trinomial, which factored into the square of a binomial.
If the problem has a degree of three (in other words the variable in the equation is cubed) then you will find three solutions. Example: .
First, factor the GCF of +2x from each of the terms of the polynomial … .
Next factor, . We now have three factors which contain variables, therefore, and and . This example of a third degree equation has three solutions .
If the equation is factored and set equal to zero as in the example, , then and .
Solve the equation, . First, you must get the equation in standard form.
This means that you must add 16 to both sides so that the –16 is removed from the left side of the equation and the equation will then be equal to zero … … . Now you are ready to factor, . Set the factors equal to zero. Since is repeated twice as a factor, there are two solutions, but they are both the same. Thus, is the only “unique” solution to this problem. This is a perfect square trinomial, which factored into the square of a binomial.
If the problem has a degree of three (in other words the variable in the equation is cubed) then you will find three solutions. Example: .
First, factor the GCF of +2x from each of the terms of the polynomial … .
Next factor, . We now have three factors which contain variables, therefore, and and . This example of a third degree equation has three solutions .
2) The square root property involves taking the square roots of both sides of an equation. Before taking the square root of each side, you must isolate the term that contains the squared variable. Once this squared-variable term is fully isolated, you will take the square root of both sides and solve for the variable. We now introduce the possibility of two roots for every square root, one positive and one negative. Place a sign in front of the side containing the constant before you take the square root of that side.
Example 1:
… the squared-variable term is isolated, so we will take the square root of
each side
each side
… notice the use of the sign, this will give us both a positive and a
negative root
… simplify both sides of the equation, here x is isolated so we have
solved this equation
negative root
… simplify both sides of the equation, here x is isolated so we have
solved this equation
Example 2:
… again the squared-variable term is isolated, so we will take the
square root of each side
… again don’t forget the sign, now simplify the radicals
square root of each side
… again don’t forget the sign, now simplify the radicals
… this time p is not fully isolated, also notice that 4 are
rational numbers, which means …
rational numbers, which means …
and
and
Example 3:
… squared term is not isolated, add 1 to each side before
beginning
beginning
… now take the square root of both sides
… simplify radicals
… radical containing the constant cannot be simplified, solve for the
variable
variable
… notice the placement of the –1 before the radical on the
right-hand side, these numbers may not be combined since
–1 is a rational number and are irrational numbers
right-hand side, these numbers may not be combined since
–1 is a rational number and are irrational numbers
In each of the first 3 examples involving the square root property, notice that there were no first-degree terms. These equations although they are quadratic in nature, have the form . To solve a quadratic equation that contains a first-degree term using the square root property would involve completing the square which is another "trick" that will be explained in another lesson.
3) The third method for solving quadratic equations described uses the quadratic formula ó Bhaskara's Formula
This formula is . If you notice, the right hand side has variables a, b, and c.
These variables are the coefficients of the terms of the quadratic equation. (Remember the standard form is .)
This formula is . If you notice, the right hand side has variables a, b, and c.
These variables are the coefficients of the terms of the quadratic equation. (Remember the standard form is .)
Example 4:
… first, the equation must be in standard form
…move terms to one side
…move terms to one side
… identify a = 1, b = –8, c = –9
… use the quadratic formula, substitute values
… simplify radical
… solve for r
and … quadratic equations have 2 solutions
RESPONDE:
Skimming:
1) ¿Cuál es el tema principal del texto?
2) ¿Cuál es la forma estándar o general de una ecuación de segundo grado?
Scanning:
3) Según el texto ¿Cuántos métodos hay para resolver ecuaciones de segundo grado?
4) Generalmente, ¿cuántas soluciones tiene una ecuación de segundo grado?
Reading in detail:
5) Utilizando la fórmula de Bhaskara resuelve: x2-x-2=0
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