Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2•32x2) -  287x) -  323  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  18x²-287x-323 

The first term is,  18x²  its coefficient is  18 .
The middle term is,  -287x  its coefficient is  -287 .
The last term, "the constant", is  -323 

Step-1 : Multiply the coefficient of the first term by the constant   18 • -323 = -5814 

Step-2 : Find two factors of  -5814  whose sum equals the coefficient of the middle term, which is   -287 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -306  and  19 
                     18x² - 306x + 19x - 323

Step-4 : Add up the first 2 terms, pulling out like factors :
                    18x • (x-17)
              Add up the last 2 terms, pulling out common factors :
                    19 • (x-17)
Step-5 : Add up the four terms of step 4 :
                    (18x+19)  •  (x-17)
             Which is the desired factorization

Equation at the end of step  2  :

  (x - 17) • (18x + 19)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    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.

Solving a Single Variable Equation :

 3.2      Solve  :    x-17 = 0 

 Add  17  to both sides of the equation : 
                      x = 17

Solving a Single Variable Equation :

 3.3      Solve  :    18x+19 = 0 

 Subtract  19  from both sides of the equation : 
                      18x = -19
Divide both sides of the equation by 18:
                     x = -19/18 = -1.056

Supplement : Solving Quadratic Equation Directly

Solving    18x2-287x-323  = 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 :

 4.1      Find the Vertex of   y = 18x²-287x-323

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, 18 , 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   7.9722  

 Plugging into the parabola formula   7.9722  for  x  we can calculate the  y -coordinate : 
  y = 18.0 * 7.97 * 7.97 - 287.0 * 7.97 - 323.0
or   y = -1467.014

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 18x²-287x-323
Axis of Symmetry (dashed)  {x}={ 7.97} 
Vertex at  {x,y} = { 7.97,-1467.01} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-1.06, 0.00} 
Root 2 at  {x,y} = {17.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   18x²-287x-323 = 0 by Completing The Square .

 Divide both sides of the equation by  18  to have 1 as the coefficient of the first term :
   x²-(287/18)x-(323/18) = 0

Add  323/18  to both side of the equation :
   x²-(287/18)x = 323/18

Now the clever bit: Take the coefficient of  x , which is  287/18 , divide by two, giving  287/36 , and finally square it giving  82369/1296 

Add  82369/1296  to both sides of the equation :
  On the right hand side we have :
   323/18  +  82369/1296   The common denominator of the two fractions is  1296   Adding  (23256/1296)+(82369/1296)  gives  105625/1296 
  So adding to both sides we finally get :
   x²-(287/18)x+(82369/1296) = 105625/1296

Adding  82369/1296  has completed the left hand side into a perfect square :
   x²-(287/18)x+(82369/1296)  =
   (x-(287/36)) • (x-(287/36))  =
  (x-(287/36))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(287/18)x+(82369/1296) = 105625/1296 and
   x²-(287/18)x+(82369/1296) = (x-(287/36))²
then, according to the law of transitivity,
   (x-(287/36))² = 105625/1296

We'll refer to this Equation as  Eq. #4.2.1  

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

Note that the square root of
   (x-(287/36))²   is
   (x-(287/36))^2/2 =
  (x-(287/36))¹ =
   x-(287/36)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   x-(287/36) = √ 105625/1296

Add  287/36  to both sides to obtain:
   x = 287/36 + √ 105625/1296

Since a square root has two values, one positive and the other negative
   x² - (287/18)x - (323/18) = 0
   has two solutions:
  x = 287/36 + √ 105625/1296
   or
  x = 287/36 - √ 105625/1296

Note that  √ 105625/1296 can be written as
  √ 105625  / √ 1296   which is 325 / 36

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    18x²-287x-323 = 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   =     18
                      B   =   -287
                      C   =  -323

Accordingly,  B²  -  4AC   =
                     82369 - (-23256) =
                     105625

Applying the quadratic formula :

               287 ± √ 105625
   x  =    ————————
                        36

Can  √ 105625 be simplified ?

Yes!   The prime factorization of  105625   is
   5•5•5•5•13•13 
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).

√ 105625   =  √ 5•5•5•5•13•13   =5•5•13•√ 1   =
                ±  325 • √ 1   =
                ±  325

So now we are looking at:
           x  =  ( 287 ± 325) / 36

Two real solutions:

x =(287+√105625)/36=(287+325)/36= 17.000

or:

x =(287-√105625)/36=(287-325)/36= -1.056

Two solutions were found :

1.  x = -19/18 = -1.056
   
2.  x = 17