2x^2 3x 14

2x^2 3x 14 DEFAULT

Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

(2x2 - 3x) - 14 = 0

Step  2  :

Trying to factor by splitting the middle term

      Factoring  2x2-3x 

The first term is,  2x2  its coefficient is  2 .
The middle term is,  -3x  its coefficient is  -3 .
The last term, "the constant", is   

Step-1 : Multiply the coefficient of the first term by the constant   2 •  =  

Step-2 : Find two factors of    whose sum equals the coefficient of the middle term, which is   -3 .

        +   1   =   
        +   2   =   
     -7   +   4   =   -3   That's it


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -7  and  4 
                     2x2 - 7x + 4x - 14

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x • (2x-7)
              Add up the last 2 terms, pulling out common factors :
                    2 • (2x-7)
Step-5 : Add up the four terms of step 4 :
                    (x+2)  •  (2x-7)
             Which is the desired factorization

Equation at the end of step  2  :

(2x - 7) • (x + 2) = 0

Step  3  :

Theory - Roots of a product :

     A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

       Solve  :    2x-7 = 0 

 Add  7  to both sides of the equation : 
                      2x = 7
Divide both sides of the equation by 2:
                     x = 7/2 =

Solving a Single Variable Equation :

       Solve  :    x+2 = 0 

 Subtract  2  from both sides of the equation : 
                      x = -2

Supplement : Solving Quadratic Equation Directly

Solving  2x2-3x  = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

       Find the Vertex of   y = 2x2-3x

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 2 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is     

 Plugging into the parabola formula     for  x  we can calculate the  y -coordinate : 
  y = * * - * -
or   y =

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 2x2-3x
Axis of Symmetry (dashed)  {x}={ } 
Vertex at  {x,y} = { ,} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {, } 
Root 2 at  {x,y} = { , } 

Solve Quadratic Equation by Completing The Square

      Solving   2x2-3x = 0 by Completing The Square .

 Divide both sides of the equation by  2  to have 1 as the coefficient of the first term :
   x2-(3/2)x-7 = 0

Add  7  to both side of the equation :
   x2-(3/2)x = 7

Now the clever bit: Take the coefficient of  x , which is  3/2 , divide by two, giving  3/4 , and finally square it giving  9/16 

Add  9/16  to both sides of the equation :
  On the right hand side we have :
   7  +  9/16    or,  (7/1)+(9/16) 
  The common denominator of the two fractions is  16   Adding  (/16)+(9/16)  gives  /16 
  So adding to both sides we finally get :
   x2-(3/2)x+(9/16) = /16

Adding  9/16  has completed the left hand side into a perfect square :
   x2-(3/2)x+(9/16)  =
   (x-(3/4)) • (x-(3/4))  =
  (x-(3/4))2
Things which are equal to the same thing are also equal to one another. Since
   x2-(3/2)x+(9/16) = /16 and
   x2-(3/2)x+(9/16) = (x-(3/4))2
then, according to the law of transitivity,
   (x-(3/4))2 = /16

We'll refer to this Equation as  Eq. #  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x-(3/4))2  is
   (x-(3/4))2/2 =
  (x-(3/4))1 =
   x-(3/4)

Now, applying the Square Root Principle to  Eq. #  we get:
   x-(3/4) = √ /16

Add  3/4  to both sides to obtain:
   x = 3/4 + √ /16

Since a square root has two values, one positive and the other negative
   x2 - (3/2)x - 7 = 0
   has two solutions:
  x = 3/4 + √ /16
   or
  x = 3/4 - √ /16

Note that  √ /16 can be written as
  √   / √ 16   which is 11 / 4

Solve Quadratic Equation using the Quadratic Formula

      Solving    2x2-3x = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B2-4AC
  x =   ————————
                      2A

  In our case,  A   =     2
                      B   =    -3
                      C   =  

Accordingly,  B2  -  4AC   =
                     9 - () =
                     

Applying the quadratic formula :

               3 ± √
   x  =    —————
                    4

Can  √ be simplified ?

Yes!   The prime factorization of     is
   11•11 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√   =  √ 11•11  =
                ±  11 • √ 1   =
                ±  11

So now we are looking at:
           x  =  ( 3 ± 11) / 4

Two real solutions:

x =(3+√)/4=(3+11)/4=

or:

x =(3-√)/4=()/4=

Two solutions were found :

  1.  x = -2
  2.  x = 7/2 =
Sours: https://www.tiger-algebra.com/drill/2x~x=0/

Simplification or other simple results

        +   -1   =           +   -2   =        -7   +   -4   =        -4   +   -7   =        -2   +      =        -1   +      =        1   +   28   =   29     2   +   14   =   16     4   +   7   =   11     7   +   4   =   11     14   +   2   =   16     28   +   1   =   29
Sours: https://www.tiger-algebra.com/drill/-2x~x/
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Solve the equation: 2x^2 + 3x =

This is a quadratic equation, for these types of questions it is often easiest to try and rearrange the formula into the form ax^2 + bx + c = 0, where a, b and c are whole numbers. Therefore the first thing to do would be to take 14 away from both sides of the equation. This would leave us with 2x^2 + 3x - 14 = 0. Now that it is in the form that we want, we can start to factorise the equation (put into brackets). I find it easiest if we focus firstly on 2x^2, the only two whole numbers that can multiply together to make 2x^2 are 2x and x (as 2 times x times x is 2x^2). This means we know what the brackets start with, (2x) and (x).
The next step would be to focus on the last number in the equation, this would be We can then write the numbers that multiply to make and 1, -7 and 2, -2 and 7, -1 and Looking at what we have in the brackets, we know that using a 14 in the equation would not produce a 3x, therefore we are left with (7 and -2) and (2 and -7). If we then try out these number in different combinations in our brackets, we will find that the only combination that produces a +3x is (2x +7)(x-2). (As (-2)2 = -4 and 7 times 1 = 7 and then 7+(-4) = 3. Therefore the factorised equation is (2x + 7)(x - 2) = 0. If this equals 0, one of the brackets must equal 0 (as 0 multiplied by anything is still 0). If x-2=0 then x=2, alternatively if 2x+7=0 then x=, therefore x = 2 or

Sours: https://www.mytutor.co.uk/answers//GCSE/Maths/Solve-the-equation-2xx/
Maya and Carina (Danielle Savre and Stefania Spampinato) Station 19 3x14 [ Marina Scenes ]
\left(x-2\right)\left(2x+7\right)
Tick mark Image
Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx To find a and b, set up a system to be solved.
a+b=3 ab=2\left(\right)=
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product
Calculate the sum for each pair.
The solution is the pair that gives sum 3.
Rewrite 2x^{2}+3x as \left(2x^{2}-4x\right)+\left(7x\right).
\left(2x^{2}-4x\right)+\left(7x\right)
Factor out 2x in the first and 7 in the second group.
2x\left(x-2\right)+7\left(x-2\right)
Factor out common term x-2 by using distributive property.
\left(x-2\right)\left(2x+7\right)
\left(x-2\right)\left(2x+7\right)
Tick mark Image
a+b=3 ab=2\left(\right)=
Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx To find a and b, set up a system to be solved.
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product
Calculate the sum for each pair.
Sours: https://mathsolver.microsoft.com/en/solve-problem/2%20x%20%5E%20%7B%%20%7D%20%2B%%20x%%

14 2x^2 3x

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Exponential equations - Exercises 1 and 2

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