Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((23•3•7v2) +  272v) -  280  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   168v² + 272v - 280  =   8 • (21v² + 34v - 35) 

Trying to factor by splitting the middle term

 3.2     Factoring  21v² + 34v - 35 

The first term is,  21v²  its coefficient is  21 .
The middle term is,  +34v  its coefficient is  34 .
The last term, "the constant", is  -35 

Step-1 : Multiply the coefficient of the first term by the constant   21 • -35 = -735 

Step-2 : Find two factors of  -735  whose sum equals the coefficient of the middle term, which is   34 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -15  and  49 
                     21v² - 15v + 49v - 35

Step-4 : Add up the first 2 terms, pulling out like factors :
                    3v • (7v-5)
              Add up the last 2 terms, pulling out common factors :
                    7 • (7v-5)
Step-5 : Add up the four terms of step 4 :
                    (3v+7)  •  (7v-5)
             Which is the desired factorization

Equation at the end of step  3  :

  8 • (7v - 5) • (3v + 7)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    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.

Equations which are never true :

 4.2      Solve :    8   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 4.3      Solve  :    7v-5 = 0 

 Add  5  to both sides of the equation : 
                      7v = 5
Divide both sides of the equation by 7:
                     v = 5/7 = 0.714

Solving a Single Variable Equation :

 4.4      Solve  :    3v+7 = 0 

 Subtract  7  from both sides of the equation : 
                      3v = -7
Divide both sides of the equation by 3:
                     v = -7/3 = -2.333

Supplement : Solving Quadratic Equation Directly

Solving    21v2+34v-35  = 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 :

 5.1      Find the Vertex of   y = 21v²+34v-35

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, 21 , 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,Av²+Bv+C,the  v -coordinate of the vertex is given by  -B/(2A) . In our case the  v  coordinate is  -0.8095  

 Plugging into the parabola formula  -0.8095  for  v  we can calculate the  y -coordinate : 
  y = 21.0 * -0.81 * -0.81 + 34.0 * -0.81 - 35.0
or   y = -48.762

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 21v²+34v-35
Axis of Symmetry (dashed)  {v}={-0.81} 
Vertex at  {v,y} = {-0.81,-48.76} 
 v -Intercepts (Roots) :
Root 1 at  {v,y} = {-2.33, 0.00} 
Root 2 at  {v,y} = { 0.71, 0.00} 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   21v²+34v-35 = 0 by Completing The Square .

 Divide both sides of the equation by  21  to have 1 as the coefficient of the first term :
   v²+(34/21)v-(5/3) = 0

Add  5/3  to both side of the equation :
   v²+(34/21)v = 5/3

Now the clever bit: Take the coefficient of  v , which is  34/21 , divide by two, giving  17/21 , and finally square it giving  289/441 

Add  289/441  to both sides of the equation :
  On the right hand side we have :
   5/3  +  289/441   The common denominator of the two fractions is  441   Adding  (735/441)+(289/441)  gives  1024/441 
  So adding to both sides we finally get :
   v²+(34/21)v+(289/441) = 1024/441

Adding  289/441  has completed the left hand side into a perfect square :
   v²+(34/21)v+(289/441)  =
   (v+(17/21)) • (v+(17/21))  =
  (v+(17/21))²
Things which are equal to the same thing are also equal to one another. Since
   v²+(34/21)v+(289/441) = 1024/441 and
   v²+(34/21)v+(289/441) = (v+(17/21))²
then, according to the law of transitivity,
   (v+(17/21))² = 1024/441

We'll refer to this Equation as  Eq. #5.2.1  

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

Note that the square root of
   (v+(17/21))²   is
   (v+(17/21))^2/2 =
  (v+(17/21))¹ =
   v+(17/21)

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:
   v+(17/21) = √ 1024/441

Subtract  17/21  from both sides to obtain:
   v = -17/21 + √ 1024/441

Since a square root has two values, one positive and the other negative
   v² + (34/21)v - (5/3) = 0
   has two solutions:
  v = -17/21 + √ 1024/441
   or
  v = -17/21 - √ 1024/441

Note that  √ 1024/441 can be written as
  √ 1024  / √ 441   which is 32 / 21

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    21v²+34v-35 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  v  , the solution for   Av²+Bv+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  v =   ————————
                      2A

  In our case,  A   =     21
                      B   =    34
                      C   =  -35

Accordingly,  B²  -  4AC   =
                     1156 - (-2940) =
                     4096

Applying the quadratic formula :

               -34 ± √ 4096
   v  =    ———————
                        42

Can  √ 4096 be simplified ?

Yes!   The prime factorization of  4096   is
   2•2•2•2•2•2•2•2•2•2•2•2 
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).

√ 4096   =  √ 2•2•2•2•2•2•2•2•2•2•2•2   =2•2•2•2•2•2•√ 1   =
                ±  64 • √ 1   =
                ±  64

So now we are looking at:
           v  =  ( -34 ± 64) / 42

Two real solutions:

v =(-34+√4096)/42=(-17+32)/21= 0.714

or:

v =(-34-√4096)/42=(-17-32)/21= -2.333

Two solutions were found :

1.  v = -7/3 = -2.333
   
2.  v = 5/7 = 0.714