Solution - Quadratic equations

Step by step solution :

Step  1  :

Trying to factor as a Difference of Squares :

 1.1      Factoring:  y²-9 

Theory : A difference of two perfect squares,  A² - B²  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A² - AB + BA - B² =
         A² - AB + AB - B² =
         A² - B²

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 9 is the square of 3
Check :  y²  is the square of  y¹ 

Factorization is :       (y + 3)  •  (y - 3) 

Equation at the end of step  1  :

  ((((y2)+y)-6)•(((y2)-6y)+9))-2•(y+3)•(y-3)  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  y²+y-6 

The first term is,  y²  its coefficient is  1 .
The middle term is,  +y  its coefficient is  1 .
The last term, "the constant", is  -6 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -6 = -6 

Step-2 : Find two factors of  -6  whose sum equals the coefficient of the middle term, which is   1 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -2  and  3 
                     y² - 2y + 3y - 6

Step-4 : Add up the first 2 terms, pulling out like factors :
                    y • (y-2)
              Add up the last 2 terms, pulling out common factors :
                    3 • (y-2)
Step-5 : Add up the four terms of step 4 :
                    (y+3)  •  (y-2)
             Which is the desired factorization

Trying to factor by splitting the middle term

 2.2     Factoring  y²-6y+9 

The first term is,  y²  its coefficient is  1 .
The middle term is,  -6y  its coefficient is  -6 .
The last term, "the constant", is  +9 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 9 = 9 

Step-2 : Find two factors of  9  whose sum equals the coefficient of the middle term, which is   -6 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -3  and  -3 
                     y² - 3y - 3y - 9

Step-4 : Add up the first 2 terms, pulling out like factors :
                    y • (y-3)
              Add up the last 2 terms, pulling out common factors :
                    3 • (y-3)
Step-5 : Add up the four terms of step 4 :
                    (y-3)  •  (y-3)
             Which is the desired factorization

Multiplying Exponential Expressions :

 2.3    Multiply  (y-3)  by  (y-3) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (y-3)  and the exponents are :
          1 , as  (y-3)  is the same number as  (y-3)¹ 
 and   1 , as  (y-3)  is the same number as  (y-3)¹ 
The product is therefore,  (y-3)⁽¹⁺¹⁾ = (y-3)² 

Equation at the end of step  2  :

  (y+3)•(y-2)•(y-3)2-2•(y+3)•(y-3)  = 0 

Step  3  :

Pulling out like terms :

 3.1      Pull out     y+3 

Pulling out like terms :

 3.2      Pull out     y-3 

After pulling out, we are left with :
      (y+3) • y-3 • ( 1  *  (y-2)  *  (y-3) +( 2  *  (-1)  *  (-1) ))

Trying to factor by splitting the middle term

 3.3     Factoring  y²-5y+8 

The first term is,  y²  its coefficient is  1 .
The middle term is,  -5y  its coefficient is  -5 .
The last term, "the constant", is  +8 

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

Step-2 : Find two factors of  8  whose sum equals the coefficient of the middle term, which is   -5 .

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

Equation at the end of step  3  :

  (y + 3) • (y - 3) • (y2 - 5y + 8)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    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 :

 4.2      Solve  :    y+3 = 0 

 Subtract  3  from both sides of the equation : 
                      y = -3

Solving a Single Variable Equation :

 4.3      Solve  :    y-3 = 0 

 Add  3  to both sides of the equation : 
                      y = 3

Parabola, Finding the Vertex :

 4.4      Find the Vertex of   t = y²-5y+8

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   2.5000  

 Plugging into the parabola formula   2.5000  for  y  we can calculate the  t -coordinate : 
  t = 1.0 * 2.50 * 2.50 - 5.0 * 2.50 + 8.0
or   t = 1.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  t = y²-5y+8
Axis of Symmetry (dashed)  {y}={ 2.50} 
Vertex at  {y,t} = { 2.50, 1.75} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 4.5     Solving   y²-5y+8 = 0 by Completing The Square .

 Subtract  8  from both side of the equation :
   y²-5y = -8

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

Add  25/4  to both sides of the equation :
  On the right hand side we have :
   -8  +  25/4    or,  (-8/1)+(25/4) 
  The common denominator of the two fractions is  4   Adding  (-32/4)+(25/4)  gives  -7/4 
  So adding to both sides we finally get :
   y²-5y+(25/4) = -7/4

Adding  25/4  has completed the left hand side into a perfect square :
   y²-5y+(25/4)  =
   (y-(5/2)) • (y-(5/2))  =
  (y-(5/2))²
Things which are equal to the same thing are also equal to one another. Since
   y²-5y+(25/4) = -7/4 and
   y²-5y+(25/4) = (y-(5/2))²
then, according to the law of transitivity,
   (y-(5/2))² = -7/4

We'll refer to this Equation as  Eq. #4.5.1  

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))²   is
   (y-(5/2))^2/2 =
  (y-(5/2))¹ =
   y-(5/2)

Now, applying the Square Root Principle to  Eq. #4.5.1  we get:
   y-(5/2) = √ -7/4

Add  5/2  to both sides to obtain:
   y = 5/2 + √ -7/4
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 - 5y + 8 = 0
   has two solutions:
  y = 5/2 + √ 7/4 •  i 
   or
  y = 5/2 - √ 7/4 •  i 

Note that  √ 7/4 can be written as
  √ 7  / √ 4   which is √ 7  / 2

Solve Quadratic Equation using the Quadratic Formula

 4.6     Solving    y²-5y+8 = 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   =    -5
                      C   =   8

Accordingly,  B²  -  4AC   =
                     25 - 32 =
                     -7

Applying the quadratic formula :

               5 ± √ -7
   y  =    —————
                    2

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,√ -7  = 
                    √ 7 • (-1)  =
                    √ 7  • √ -1   =
                    ±  √ 7  • i

  √ 7   , rounded to 4 decimal digits, is   2.6458
 So now we are looking at:
           y  =  ( 5 ±  2.646 i ) / 2

Two imaginary solutions :

 y =(5+√-7)/2=(5+i√ 7 )/2= 2.5000+1.3229i
  or: 
 y =(5-√-7)/2=(5-i√ 7 )/2= 2.5000-1.3229i

Supplement : Solving Quadratic Equation Directly

Solving    y2-6y+9  = 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 :

 5.1      Find the Vertex of   t = y²-6y+9

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   3.0000  

 Plugging into the parabola formula   3.0000  for  y  we can calculate the  t -coordinate : 
  t = 1.0 * 3.00 * 3.00 - 6.0 * 3.00 + 9.0
or   t = 0.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  t = y²-6y+9
Vertex at  {y,t} = { 3.00, 0.00} 
y-Intercept (Root) :
One Root at  {y,t}={ 3.00, 0.00} 
Note that the root coincides with
the Vertex and the Axis of Symmetry
coinsides with the line  y = 0 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   y²-6y+9 = 0 by Completing The Square .

 Subtract  9  from both side of the equation :
   y²-6y = -9

Now the clever bit: Take the coefficient of  y , which is  6 , divide by two, giving  3 , and finally square it giving  9 

Add  9  to both sides of the equation :
  On the right hand side we have :
   -9  +  9    or,  (-9/1)+(9/1) 
  The common denominator of the two fractions is  1   Adding  (-9/1)+(9/1)  gives  0/1 
  So adding to both sides we finally get :
   y²-6y+9 = 0

Adding  9  has completed the left hand side into a perfect square :
   y²-6y+9  =
   (y-3) • (y-3)  =
  (y-3)²
Things which are equal to the same thing are also equal to one another. Since
   y²-6y+9 = 0 and
   y²-6y+9 = (y-3)²
then, according to the law of transitivity,
   (y-3)² = 0

We'll refer to this Equation as  Eq. #5.2.1  

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

Note that the square root of
   (y-3)²   is
   (y-3)^2/2 =
  (y-3)¹ =
   y-3

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:
   y-3 = √ 0

Add  3  to both sides to obtain:
   y = 3 + √ 0
The square root of zero is zero

This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.

The solution is:
   y  =  3 

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    y²-6y+9 = 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   =    -6
                      C   =   9

Accordingly,  B²  -  4AC   =
                     36 - 36 =
                     0

Applying the quadratic formula :

               6 ± √ 0
   y  =    ————
                   2

The square root of zero is zero

This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.

The solution is:
  y = 6 / 2 = 3

Four solutions were found :

1.  y =(5-√-7)/2=(5-i√ 7 )/2= 2.5000-1.3229i
2.  y =(5+√-7)/2=(5+i√ 7 )/2= 2.5000+1.3229i
3.  y = 3
4.  y = -3