Solution - Quadratic equations

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "^-3" was replaced by "^(-3)".

Rearrange:

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

                     y^(-3)-(8/125)=0 

Step by step solution :

Step  1  :

             8 
 Simplify   ———
            125

Equation at the end of step  1  :

            8 
  (y-3) -  ———  = 0 
           125
 Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  125  as the denominator :

              y(-3)      y(-3) • 125 
     y(-3) =  —————  =  ———————————
                1           125    

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 :

 2.2       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:

 y(-3) • 125 - (8)      125y(-3) - 8 
 —————————————————  =  ————————————
        125                125     

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   125y^(-3) - 8  =   -y^(-3) • (8y³ - 125) 

Trying to factor as a Difference of Cubes:

 3.2      Factoring:  8y³ - 125 

Theory : A difference of two perfect cubes,  a³ - b³ can be factored into
              (a-b) • (a² +ab +b²)

Proof :  (a-b)•(a²+ab+b²) =
            a³+a²b+ab²-ba²-b²a-b³ =
            a³+(a²b-ba²)+(ab²-b²a)-b³ =
            a³+0+0-b³ =
            a³-b³

Check :  8  is the cube of  2 

Check :  125  is the cube of   5 
Check :  y³ is the cube of   y¹

Factorization is :
             (2y - 5)  •  (4y² + 10y + 25) 

Trying to factor by splitting the middle term

 3.3     Factoring  4y² + 10y + 25 

The first term is,  4y²  its coefficient is  4 .
The middle term is,  +10y  its coefficient is  10 .
The last term, "the constant", is  +25 

Step-1 : Multiply the coefficient of the first term by the constant   4 • 25 = 100 

Step-2 : Find two factors of  100  whose sum equals the coefficient of the middle term, which is   10 .

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  3  :

  -y(-3) • (2y - 5) • (4y2 + 10y + 25) 
  ————————————————————————————————————  = 0 
                  125                 

Step  4  :

When a fraction equals zero :

 4.1    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:

  -y(-3)•(2y-5)•(4y2+10y+25) 
  —————————————————————————— • 125 = 0 • 125
             125            

Now, on the left hand side, the  125  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -y^(-3)  •  (2y-5)  •  (4y²+10y+25)  = 0

Theory - Roots of a product :

 4.2    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 :

 4.3      Solve  :    -y^(-3) = 0 

  This equation has no solution !!

We actually looking at 1/

Solving a Single Variable Equation :

 4.4      Solve  :    2y-5 = 0 

 Add  5  to both sides of the equation : 
                      2y = 5
Divide both sides of the equation by 2:
                     y = 5/2 = 2.500

Parabola, Finding the Vertex :

 4.5      Find the Vertex of   t = 4y²+10y+25

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, 4 , 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  -1.2500  

 Plugging into the parabola formula  -1.2500  for  y  we can calculate the  t -coordinate : 
  t = 4.0 * -1.25 * -1.25 + 10.0 * -1.25 + 25.0
or   t = 18.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  t = 4y²+10y+25
Axis of Symmetry (dashed)  {y}={-1.25} 
Vertex at  {y,t} = {-1.25,18.75} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 4.6     Solving   4y²+10y+25 = 0 by Completing The Square .

 Divide both sides of the equation by  4  to have 1 as the coefficient of the first term :
   y²+(5/2)y+(25/4) = 0

Subtract  25/4  from both side of the equation :
   y²+(5/2)y = -25/4

Now the clever bit: Take the coefficient of  y , which is  5/2 , divide by two, giving  5/4 , and finally square it giving  25/16 

Add  25/16  to both sides of the equation :
  On the right hand side we have :
   -25/4  +  25/16   The common denominator of the two fractions is  16   Adding  (-100/16)+(25/16)  gives  -75/16 
  So adding to both sides we finally get :
   y²+(5/2)y+(25/16) = -75/16

Adding  25/16  has completed the left hand side into a perfect square :
   y²+(5/2)y+(25/16)  =
   (y+(5/4)) • (y+(5/4))  =
  (y+(5/4))²
Things which are equal to the same thing are also equal to one another. Since
   y²+(5/2)y+(25/16) = -75/16 and
   y²+(5/2)y+(25/16) = (y+(5/4))²
then, according to the law of transitivity,
   (y+(5/4))² = -75/16

We'll refer to this Equation as  Eq. #4.6.1  

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

Note that the square root of
   (y+(5/4))²   is
   (y+(5/4))^2/2 =
  (y+(5/4))¹ =
   y+(5/4)

Now, applying the Square Root Principle to  Eq. #4.6.1  we get:
   y+(5/4) = √ -75/16

Subtract  5/4  from both sides to obtain:
   y = -5/4 + √ -75/16
In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 


Since a square root has two values, one positive and the other negative
   y² + (5/2)y + (25/4) = 0
   has two solutions:
  y = -5/4 + √ 75/16 •  i 
   or
  y = -5/4 - √ 75/16 •  i 

Note that  √ 75/16 can be written as
  √ 75  / √ 16   which is √ 75  / 4

Solve Quadratic Equation using the Quadratic Formula

 4.7     Solving    4y²+10y+25 = 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   =     4
                      B   =    10
                      C   =   25

Accordingly,  B²  -  4AC   =
                     100 - 400 =
                     -300

Applying the quadratic formula :

               -10 ± √ -300
   y  =    ——————
                      8

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -300  = 
                    √ 300 • (-1)  =
                    √ 300  • √ -1   =
                    ±  √ 300  • i

Can  √ 300 be simplified ?

Yes!   The prime factorization of  300   is
   2•2•3•5•5 
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).

√ 300   =  √ 2•2•3•5•5   =2•5•√ 3   =
                ±  10 • √ 3

  √ 3   , rounded to 4 decimal digits, is   1.7321
 So now we are looking at:
           y  =  ( -10 ± 10 •  1.732 i ) / 8

Two imaginary solutions :

 y =(-10+√-300)/8=(-5+5i√ 3 )/4= -1.2500+2.1651i
  or: 
 y =(-10-√-300)/8=(-5-5i√ 3 )/4= -1.2500-2.1651i

Three solutions were found :

1.  y =(-10-√-300)/8=(-5-5i√ 3 )/4= -1.2500-2.1651i
2.  y =(-10+√-300)/8=(-5+5i√ 3 )/4= -1.2500+2.1651i
3.  y = 5/2 = 2.500