Solution - Quadratic equations

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

 1.1     Factoring  y²-10y-7 

The first term is,  y²  its coefficient is  1 .
The middle term is,  -10y  its coefficient is  -10 .
The last term, "the constant", is  -7 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -7 = -7 

Step-2 : Find two factors of  -7  whose sum equals the coefficient of the middle term, which is   -10 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  1  :

  y2 - 10y - 7  = 0 

Step  2  :

Parabola, Finding the Vertex :

 2.1      Find the Vertex of   t = y²-10y-7

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  "t"  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,Ay²+By+C,the  y -coordinate of the vertex is given by  -B/(2A) . In our case the  y  coordinate is   5.0000  

 Plugging into the parabola formula   5.0000  for  y  we can calculate the  t -coordinate : 
  t = 1.0 * 5.00 * 5.00 - 10.0 * 5.00 - 7.0
or   t = -32.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  t = y²-10y-7
Axis of Symmetry (dashed)  {y}={ 5.00} 
Vertex at  {y,t} = { 5.00,-32.00} 
 y -Intercepts (Roots) :
Root 1 at  {y,t} = {-0.66, 0.00} 
Root 2 at  {y,t} = {10.66, 0.00} 

Solve Quadratic Equation by Completing The Square

 2.2     Solving   y²-10y-7 = 0 by Completing The Square .

 Add  7  to both side of the equation :
   y²-10y = 7

Now the clever bit: Take the coefficient of  y , which is  10 , divide by two, giving  5 , and finally square it giving  25 

Add  25  to both sides of the equation :
  On the right hand side we have :
   7  +  25    or,  (7/1)+(25/1) 
  The common denominator of the two fractions is  1   Adding  (7/1)+(25/1)  gives  32/1 
  So adding to both sides we finally get :
   y²-10y+25 = 32

Adding  25  has completed the left hand side into a perfect square :
   y²-10y+25  =
   (y-5) • (y-5)  =
  (y-5)²
Things which are equal to the same thing are also equal to one another. Since
   y²-10y+25 = 32 and
   y²-10y+25 = (y-5)²
then, according to the law of transitivity,
   (y-5)² = 32

We'll refer to this Equation as  Eq. #2.2.1  

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

Note that the square root of
   (y-5)²   is
   (y-5)^2/2 =
  (y-5)¹ =
   y-5

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
   y-5 = √ 32

Add  5  to both sides to obtain:
   y = 5 + √ 32

Since a square root has two values, one positive and the other negative
   y² - 10y - 7 = 0
   has two solutions:
  y = 5 + √ 32
   or
  y = 5 - √ 32

Solve Quadratic Equation using the Quadratic Formula

 2.3     Solving    y²-10y-7 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  y  , the solution for   Ay²+By+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  y =   ————————
                      2A

  In our case,  A   =     1
                      B   =   -10
                      C   =   -7

Accordingly,  B²  -  4AC   =
                     100 - (-28) =
                     128

Applying the quadratic formula :

               10 ± √ 128
   y  =    —————
                    2

Can  √ 128 be simplified ?

Yes!   The prime factorization of  128   is
   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).

√ 128   =  √ 2•2•2•2•2•2•2   =2•2•2•√ 2   =
                ±  8 • √ 2

  √ 2   , rounded to 4 decimal digits, is   1.4142
 So now we are looking at:
           y  =  ( 10 ± 8 •  1.414 ) / 2

Two real solutions:

 y =(10+√128)/2=5+4√ 2 = 10.657

or:

 y =(10-√128)/2=5-4√ 2 = -0.657

Two solutions were found :

1.  y =(10-√128)/2=5-4√ 2 = -0.657
2.  y =(10+√128)/2=5+4√ 2 = 10.657