Solution - Quadratic equations

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "q2"   was replaced by   "q^2". 

Rearrange:

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

                     -8*q^2-(-q-1)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (0 -  23q2) -  (-q - 1)  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  -8q²+q+1 

The first term is,  -8q²  its coefficient is  -8 .
The middle term is,  +q  its coefficient is  1 .
The last term, "the constant", is  +1 

Step-1 : Multiply the coefficient of the first term by the constant   -8 • 1 = -8 

Step-2 : Find two factors of  -8  whose sum equals the coefficient of the middle term, which is   1 .

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

Equation at the end of step  2  :

  -8q2 + q + 1  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = -8q²+q+1

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -8 , is negative (smaller 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,Aq²+Bq+C,the  q -coordinate of the vertex is given by  -B/(2A) . In our case the  q  coordinate is   0.0625  

 Plugging into the parabola formula   0.0625  for  q  we can calculate the  y -coordinate : 
  y = -8.0 * 0.06 * 0.06 + 1.0 * 0.06 + 1.0
or   y = 1.031

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -8q²+q+1
Axis of Symmetry (dashed)  {q}={ 0.06} 
Vertex at  {q,y} = { 0.06, 1.03} 
 q -Intercepts (Roots) :
Root 1 at  {q,y} = { 0.42, 0.00} 
Root 2 at  {q,y} = {-0.30, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   -8q²+q+1 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 8q²-q-1 = 0  Divide both sides of the equation by  8  to have 1 as the coefficient of the first term :
   q²-(1/8)q-(1/8) = 0

Add  1/8  to both side of the equation :
   q²-(1/8)q = 1/8

Now the clever bit: Take the coefficient of  q , which is  1/8 , divide by two, giving  1/16 , and finally square it giving  1/256 

Add  1/256  to both sides of the equation :
  On the right hand side we have :
   1/8  +  1/256   The common denominator of the two fractions is  256   Adding  (32/256)+(1/256)  gives  33/256 
  So adding to both sides we finally get :
   q²-(1/8)q+(1/256) = 33/256

Adding  1/256  has completed the left hand side into a perfect square :
   q²-(1/8)q+(1/256)  =
   (q-(1/16)) • (q-(1/16))  =
  (q-(1/16))²
Things which are equal to the same thing are also equal to one another. Since
   q²-(1/8)q+(1/256) = 33/256 and
   q²-(1/8)q+(1/256) = (q-(1/16))²
then, according to the law of transitivity,
   (q-(1/16))² = 33/256

We'll refer to this Equation as  Eq. #3.2.1  

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

Note that the square root of
   (q-(1/16))²   is
   (q-(1/16))^2/2 =
  (q-(1/16))¹ =
   q-(1/16)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   q-(1/16) = √ 33/256

Add  1/16  to both sides to obtain:
   q = 1/16 + √ 33/256

Since a square root has two values, one positive and the other negative
   q² - (1/8)q - (1/8) = 0
   has two solutions:
  q = 1/16 + √ 33/256
   or
  q = 1/16 - √ 33/256

Note that  √ 33/256 can be written as
  √ 33  / √ 256   which is √ 33  / 16

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    -8q²+q+1 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  q  , the solution for   Aq²+Bq+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  q =   ————————
                      2A

  In our case,  A   =     -8
                      B   =    1
                      C   =   1

Accordingly,  B²  -  4AC   =
                     1 - (-32) =
                     33

Applying the quadratic formula :

               -1 ± √ 33
   q  =    —————
                    -16

  √ 33   , rounded to 4 decimal digits, is   5.7446
 So now we are looking at:
           q  =  ( -1 ±  5.745 ) / -16

Two real solutions:

 q =(-1+√33)/-16=-0.297

or:

 q =(-1-√33)/-16= 0.422

Two solutions were found :

1.  q =(-1-√33)/-16= 0.422
2.  q =(-1+√33)/-16=-0.297