Solution - Quadratic equations

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "0.25" was replaced by "(25/100)".

Step by step solution :

Step  1  :

            1
 Simplify   —
            4

Equation at the end of step  1  :

    1                  
  ((— • x2) -  7x) -  15  = 0 
    4                  

Step  2  :

Equation at the end of step  2  :

   x2           
  (—— -  7x) -  15  = 0 
   4            

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  4  as the denominator :

          7x     7x • 4
    7x =  ——  =  ——————
          1        4   

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 3.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 x2 - (7x • 4)     x2 - 28x
 —————————————  =  ————————
       4              4    

Equation at the end of step  3  :

  (x2 - 28x)    
  —————————— -  15  = 0 
      4         

Step  4  :

Rewriting the whole as an Equivalent Fraction :

 4.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  4  as the denominator :

          15     15 • 4
    15 =  ——  =  ——————
          1        4   

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   x² - 28x  =   x • (x - 28) 

Adding fractions that have a common denominator :

 5.2       Adding up the two equivalent fractions

 x • (x-28) - (15 • 4)     x2 - 28x - 60
 —————————————————————  =  —————————————
           4                     4      

Trying to factor by splitting the middle term

 5.3     Factoring  x² - 28x - 60 

The first term is,  x²  its coefficient is  1 .
The middle term is,  -28x  its coefficient is  -28 .
The last term, "the constant", is  -60 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -60 = -60 

Step-2 : Find two factors of  -60  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,  -30  and  2 
                     x² - 30x + 2x - 60

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 :
                    2 • (x-30)
Step-5 : Add up the four terms of step 4 :
                    (x+2)  •  (x-30)
             Which is the desired factorization

Equation at the end of step  5  :

  (x + 2) • (x - 30)
  ——————————————————  = 0 
          4         

Step  6  :

When a fraction equals zero :

 6.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  (x+2)•(x-30)
  ———————————— • 4 = 0 • 4
       4      

Now, on the left hand side, the  4  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (x+2)  •  (x-30)  = 0

Theory - Roots of a product :

 6.2    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 :

 6.3      Solve  :    x+2 = 0 

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

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    x2-28x-60  = 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 = x²-28x-60

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  14.0000  

 Plugging into the parabola formula  14.0000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * 14.00 * 14.00 - 28.0 * 14.00 - 60.0
or   y = -256.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²-28x-60
Axis of Symmetry (dashed)  {x}={14.00} 
Vertex at  {x,y} = {14.00,-256.00} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-2.00, 0.00} 
Root 2 at  {x,y} = {30.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 7.2     Solving   x²-28x-60 = 0 by Completing The Square .

 Add  60  to both side of the equation :
   x²-28x = 60

Now the clever bit: Take the coefficient of  x , 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 :
   60  +  196    or,  (60/1)+(196/1) 
  The common denominator of the two fractions is  1   Adding  (60/1)+(196/1)  gives  256/1 
  So adding to both sides we finally get :
   x²-28x+196 = 256

Adding  196  has completed the left hand side into a perfect square :
   x²-28x+196  =
   (x-14) • (x-14)  =
  (x-14)²
Things which are equal to the same thing are also equal to one another. Since
   x²-28x+196 = 256 and
   x²-28x+196 = (x-14)²
then, according to the law of transitivity,
   (x-14)² = 256

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-14)²   is
   (x-14)^2/2 =
  (x-14)¹ =
   x-14

Now, applying the Square Root Principle to  Eq. #7.2.1  we get:
   x-14 = √ 256

Add  14  to both sides to obtain:
   x = 14 + √ 256

Since a square root has two values, one positive and the other negative
   x² - 28x - 60 = 0
   has two solutions:
  x = 14 + √ 256
   or
  x = 14 - √ 256

Solve Quadratic Equation using the Quadratic Formula

 7.3     Solving    x²-28x-60 = 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   =   -28
                      C   =  -60

Accordingly,  B²  -  4AC   =
                     784 - (-240) =
                     1024

Applying the quadratic formula :

               28 ± √ 1024
   x  =    ——————
                      2

Can  √ 1024 be simplified ?

Yes!   The prime factorization of  1024   is
   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).

√ 1024   =  √ 2•2•2•2•2•2•2•2•2•2   =2•2•2•2•2•√ 1   =
                ±  32 • √ 1   =
                ±  32

So now we are looking at:
           x  =  ( 28 ± 32) / 2

Two real solutions:

x =(28+√1024)/2=14+16= 30.000

or:

x =(28-√1024)/2=14-16= -2.000

Two solutions were found :

1.  x = 30
2.  x = -2