Solution - Quadratic equations

Rearrange:

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

                (40-2*x)*(30-2*x)-(600)=0 

Step by step solution :

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   40 - 2x  =   -2 • (x - 20) 

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   (30 - 2x)  =   -2 • (x - 15) 

Equation at the end of step  3  :

  4 • (x - 20) • (x - 15) -  600  = 0 

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   4x² - 140x + 600  =   4 • (x² - 35x + 150) 

Trying to factor by splitting the middle term

 5.2     Factoring  x² - 35x + 150 

The first term is,  x²  its coefficient is  1 .
The middle term is,  -35x  its coefficient is  -35 .
The last term, "the constant", is  +150 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 150 = 150 

Step-2 : Find two factors of  150  whose sum equals the coefficient of the middle term, which is   -35 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -30  and  -5 
                     x² - 30x - 5x - 150

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x • (x-30)
              Add up the last 2 terms, pulling out common factors :
                    5 • (x-30)
Step-5 : Add up the four terms of step 4 :
                    (x-5)  •  (x-30)
             Which is the desired factorization

Equation at the end of step  5  :

  4 • (x - 5) • (x - 30)  = 0 

Step  6  :

Theory - Roots of a product :

 6.1    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 :

 6.2      Solve :    4   =  0

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

Solving a Single Variable Equation :

 6.3      Solve  :    x-5 = 0 

 Add  5  to both sides of the equation : 
                      x = 5

Solving a Single Variable Equation :

 6.4      Solve  :    x-30 = 0 

 Add  30  to both sides of the equation : 
                      x = 30

Supplement : Solving Quadratic Equation Directly

Solving    x2-35x+150  = 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 :

 7.1      Find the Vertex of   y = x²-35x+150

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,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  17.5000  

 Plugging into the parabola formula  17.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * 17.50 * 17.50 - 35.0 * 17.50 + 150.0
or   y = -156.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²-35x+150
Axis of Symmetry (dashed)  {x}={17.50} 
Vertex at  {x,y} = {17.50,-156.25} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 5.00, 0.00} 
Root 2 at  {x,y} = {30.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 7.2     Solving   x²-35x+150 = 0 by Completing The Square .

 Subtract  150  from both side of the equation :
   x²-35x = -150

Now the clever bit: Take the coefficient of  x , which is  35 , divide by two, giving  35/2 , and finally square it giving  1225/4 

Add  1225/4  to both sides of the equation :
  On the right hand side we have :
   -150  +  1225/4    or,  (-150/1)+(1225/4) 
  The common denominator of the two fractions is  4   Adding  (-600/4)+(1225/4)  gives  625/4 
  So adding to both sides we finally get :
   x²-35x+(1225/4) = 625/4

Adding  1225/4  has completed the left hand side into a perfect square :
   x²-35x+(1225/4)  =
   (x-(35/2)) • (x-(35/2))  =
  (x-(35/2))²
Things which are equal to the same thing are also equal to one another. Since
   x²-35x+(1225/4) = 625/4 and
   x²-35x+(1225/4) = (x-(35/2))²
then, according to the law of transitivity,
   (x-(35/2))² = 625/4

We'll refer to this Equation as  Eq. #7.2.1  

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

Note that the square root of
   (x-(35/2))²   is
   (x-(35/2))^2/2 =
  (x-(35/2))¹ =
   x-(35/2)

Now, applying the Square Root Principle to  Eq. #7.2.1  we get:
   x-(35/2) = √ 625/4

Add  35/2  to both sides to obtain:
   x = 35/2 + √ 625/4

Since a square root has two values, one positive and the other negative
   x² - 35x + 150 = 0
   has two solutions:
  x = 35/2 + √ 625/4
   or
  x = 35/2 - √ 625/4

Note that  √ 625/4 can be written as
  √ 625  / √ 4   which is 25 / 2

Solve Quadratic Equation using the Quadratic Formula

 7.3     Solving    x²-35x+150 = 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   =     1
                      B   =   -35
                      C   =  150

Accordingly,  B²  -  4AC   =
                     1225 - 600 =
                     625

Applying the quadratic formula :

               35 ± √ 625
   x  =    ——————
                      2

Can  √ 625 be simplified ?

Yes!   The prime factorization of  625   is
   5•5•5•5 
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).

√ 625   =  √ 5•5•5•5   =5•5•√ 1   =
                ±  25 • √ 1   =
                ±  25

So now we are looking at:
           x  =  ( 35 ± 25) / 2

Two real solutions:

x =(35+√625)/2=(35+25)/2= 30.000

or:

x =(35-√625)/2=(35-25)/2= 5.000

Two solutions were found :

1.  x = 30
2.  x = 5