Solution - Quadratic equations

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     h^2+(h+28)^2-(52^2)=0 

Step by step solution :

Step  1  :

 1.1     52 = 22•13

(52)² = (2²•13)² = 2⁴ • 13²

Equation at the end of step  1  :

  ((h2) +  ((h + 28)2)) -  (24•132)  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2h² + 56h - 1920  =   2 • (h² + 28h - 960) 

Trying to factor by splitting the middle term

 3.2     Factoring  h² + 28h - 960 

The first term is,  h²  its coefficient is  1 .
The middle term is,  +28h  its coefficient is  28 .
The last term, "the constant", is  -960 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -960 = -960 

Step-2 : Find two factors of  -960  whose sum equals the coefficient of the middle term, which is   28 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -20  and  48 
                     h² - 20h + 48h - 960

Step-4 : Add up the first 2 terms, pulling out like factors :
                    h • (h-20)
              Add up the last 2 terms, pulling out common factors :
                    48 • (h-20)
Step-5 : Add up the four terms of step 4 :
                    (h+48)  •  (h-20)
             Which is the desired factorization

Equation at the end of step  3  :

  2 • (h + 48) • (h - 20)  = 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 :    2   =  0

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

Solving a Single Variable Equation :

 4.3      Solve  :    h+48 = 0 

 Subtract  48  from both sides of the equation : 
                      h = -48

Solving a Single Variable Equation :

 4.4      Solve  :    h-20 = 0 

 Add  20  to both sides of the equation : 
                      h = 20

Supplement : Solving Quadratic Equation Directly

Solving    h2+28h-960  = 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 = h²+28h-960

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, 1 , 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,Ah²+Bh+C,the  h -coordinate of the vertex is given by  -B/(2A) . In our case the  h  coordinate is  -14.0000  

 Plugging into the parabola formula  -14.0000  for  h  we can calculate the  y -coordinate : 
  y = 1.0 * -14.00 * -14.00 + 28.0 * -14.00 - 960.0
or   y = -1156.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = h²+28h-960
Axis of Symmetry (dashed)  {h}={-14.00} 
Vertex at  {h,y} = {-14.00,-1156.00} 
 h -Intercepts (Roots) :
Root 1 at  {h,y} = {-48.00, 0.00} 
Root 2 at  {h,y} = {20.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   h²+28h-960 = 0 by Completing The Square .

 Add  960  to both side of the equation :
   h²+28h = 960

Now the clever bit: Take the coefficient of  h , which is  28 , divide by two, giving  14 , and finally square it giving  196 

Add  196  to both sides of the equation :
  On the right hand side we have :
   960  +  196    or,  (960/1)+(196/1) 
  The common denominator of the two fractions is  1   Adding  (960/1)+(196/1)  gives  1156/1 
  So adding to both sides we finally get :
   h²+28h+196 = 1156

Adding  196  has completed the left hand side into a perfect square :
   h²+28h+196  =
   (h+14) • (h+14)  =
  (h+14)²
Things which are equal to the same thing are also equal to one another. Since
   h²+28h+196 = 1156 and
   h²+28h+196 = (h+14)²
then, according to the law of transitivity,
   (h+14)² = 1156

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
   (h+14)²   is
   (h+14)^2/2 =
  (h+14)¹ =
   h+14

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:
   h+14 = √ 1156

Subtract  14  from both sides to obtain:
   h = -14 + √ 1156

Since a square root has two values, one positive and the other negative
   h² + 28h - 960 = 0
   has two solutions:
  h = -14 + √ 1156
   or
  h = -14 - √ 1156

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    h²+28h-960 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  h  , the solution for   Ah²+Bh+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  h =   ————————
                      2A

  In our case,  A   =     1
                      B   =    28
                      C   =  -960

Accordingly,  B²  -  4AC   =
                     784 - (-3840) =
                     4624

Applying the quadratic formula :

               -28 ± √ 4624
   h  =    ———————
                        2

Can  √ 4624 be simplified ?

Yes!   The prime factorization of  4624   is
   2•2•2•2•17•17 
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).

√ 4624   =  √ 2•2•2•2•17•17   =2•2•17•√ 1   =
                ±  68 • √ 1   =
                ±  68

So now we are looking at:
           h  =  ( -28 ± 68) / 2

Two real solutions:

h =(-28+√4624)/2=-14+34= 20.000

or:

h =(-28-√4624)/2=-14-34= -48.000

Two solutions were found :

1.  h = 20
2.  h = -48