Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "0.43" was replaced by "(43/100)". 2 more similar replacement(s)

Step by step solution :

Step  1  :

             43
 Simplify   ———
            100

Equation at the end of step  1  :

            151           43
  ((q2) +  (——— • q)) -  ———  = 0 
            100          100

Step  2  :

            151
 Simplify   ———
            100

Equation at the end of step  2  :

            151           43
  ((q2) +  (——— • q)) -  ———  = 0 
            100          100
 Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  100  as the denominator :

           q2     q2 • 100
     q2 =  ——  =  ————————
           1        100   

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:

 q2 • 100 + 151q     100q2 + 151q
 ———————————————  =  ————————————
       100               100     

Equation at the end of step  3  :

  (100q2 + 151q)     43
  —————————————— -  ———  = 0 
       100          100

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   100q² + 151q  =   q • (100q + 151) 

Adding fractions which have a common denominator :

 5.2       Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 q • (100q+151) - (43)     100q2 + 151q - 43
 —————————————————————  =  —————————————————
          100                     100       

Trying to factor by splitting the middle term

 5.3     Factoring  100q² + 151q - 43 

The first term is,  100q²  its coefficient is  100 .
The middle term is,  +151q  its coefficient is  151 .
The last term, "the constant", is  -43 

Step-1 : Multiply the coefficient of the first term by the constant   100 • -43 = -4300 

Step-2 : Find two factors of  -4300  whose sum equals the coefficient of the middle term, which is   151 .

For tidiness, printing of 12 lines which failed to find two such factors, was suppressed

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

Equation at the end of step  5  :

  100q2 + 151q - 43
  —————————————————  = 0 
         100       

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:

  100q2+151q-43
  ————————————— • 100 = 0 • 100
       100     

Now, on the left hand side, the  100  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   100q²+151q-43  = 0

Parabola, Finding the Vertex :

 6.2      Find the Vertex of   y = 100q²+151q-43

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, 100 , 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,Aq²+Bq+C,the  q -coordinate of the vertex is given by  -B/(2A) . In our case the  q  coordinate is  -0.7550  

 Plugging into the parabola formula  -0.7550  for  q  we can calculate the  y -coordinate : 
  y = 100.0 * -0.76 * -0.76 + 151.0 * -0.76 - 43.0
or   y = -100.002

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 100q²+151q-43
Axis of Symmetry (dashed)  {q}={-0.76} 
Vertex at  {q,y} = {-0.76,-100.00} 
 q -Intercepts (Roots) :
Root 1 at  {q,y} = {-1.76, 0.00} 
Root 2 at  {q,y} = { 0.25, 0.00} 

Solve Quadratic Equation by Completing The Square

 6.3     Solving   100q²+151q-43 = 0 by Completing The Square .

 Divide both sides of the equation by  100  to have 1 as the coefficient of the first term :
   q²+(151/100)q-(43/100) = 0

Add  43/100  to both side of the equation :
   q²+(151/100)q = 43/100

Now the clever bit: Take the coefficient of  q , which is  151/100 , divide by two, giving  151/200 , and finally square it giving  22801/40000 

Add  22801/40000  to both sides of the equation :
  On the right hand side we have :
   43/100  +  22801/40000   The common denominator of the two fractions is  40000   Adding  (17200/40000)+(22801/40000)  gives  40001/40000 
  So adding to both sides we finally get :
   q²+(151/100)q+(22801/40000) = 40001/40000

Adding  22801/40000  has completed the left hand side into a perfect square :
   q²+(151/100)q+(22801/40000)  =
   (q+(151/200)) • (q+(151/200))  =
  (q+(151/200))²
Things which are equal to the same thing are also equal to one another. Since
   q²+(151/100)q+(22801/40000) = 40001/40000 and
   q²+(151/100)q+(22801/40000) = (q+(151/200))²
then, according to the law of transitivity,
   (q+(151/200))² = 40001/40000

We'll refer to this Equation as  Eq. #6.3.1  

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

Note that the square root of
   (q+(151/200))²   is
   (q+(151/200))^2/2 =
  (q+(151/200))¹ =
   q+(151/200)

Now, applying the Square Root Principle to  Eq. #6.3.1  we get:
   q+(151/200) = √ 40001/40000

Subtract  151/200  from both sides to obtain:
   q = -151/200 + √ 40001/40000

Since a square root has two values, one positive and the other negative
   q² + (151/100)q - (43/100) = 0
   has two solutions:
  q = -151/200 + √ 40001/40000
   or
  q = -151/200 - √ 40001/40000

Note that  √ 40001/40000 can be written as
  √ 40001  / √ 40000   which is √ 40001  / 200

Solve Quadratic Equation using the Quadratic Formula

 6.4     Solving    100q²+151q-43 = 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   =    100
                      B   =   151
                      C   =  -43

Accordingly,  B²  -  4AC   =
                     22801 - (-17200) =
                     40001

Applying the quadratic formula :

               -151 ± √ 40001
   q  =    ————————
                        200

  √ 40001   , rounded to 4 decimal digits, is  200.0025
 So now we are looking at:
           q  =  ( -151 ±  200.002 ) / 200

Two real solutions:

 q =(-151+√40001)/200= 0.245

or:

 q =(-151-√40001)/200=-1.755

Two solutions were found :

1.  q =(-151-√40001)/200=-1.755
2.  q =(-151+√40001)/200= 0.245