Solution - Equations reducible to quadratic form

Rearrange:

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

                     7/x^2-(8-x^2)=0 

Step by step solution :

Step  1  :

             7
 Simplify   ——
            x2

Equation at the end of step  1  :

   7    
  —— -  (8 - x2)  = 0 
  x2    

Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  x²  as the denominator :

              8 - x2     (8 - x2) • x2
    8 - x2 =  ——————  =  —————————————
                1             x2      

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

Trying to factor as a Difference of Squares :

 2.2      Factoring:  8 - x² 

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 :  8  is not a square !!

Ruling : Binomial can not be factored as the
difference of two perfect squares

Adding fractions that have a common denominator :

 2.3       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:

 7 - ((8-x2) • x2)     x4 - 8x2 + 7
 —————————————————  =  ————————————
        x2                  x2     

Trying to factor by splitting the middle term

 2.4     Factoring  x⁴ - 8x² + 7 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  -8x²  its coefficient is  -8 .
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   -8 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -7  and  -1 
                     x⁴ - 7x² - 1x² - 7

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x² • (x²-7)
              Add up the last 2 terms, pulling out common factors :
                     1 • (x²-7)
Step-5 : Add up the four terms of step 4 :
                    (x²-1)  •  (x²-7)
             Which is the desired factorization

Trying to factor as a Difference of Squares :

 2.5      Factoring:  x²-1 

Check : 1 is the square of 1
Check :  x²  is the square of  x¹ 

Factorization is :       (x + 1)  •  (x - 1) 

Trying to factor as a Difference of Squares :

 2.6      Factoring:  x² - 7 

Check : 7 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step  2  :

  (x + 1) • (x - 1) • (x2 - 7)
  ————————————————————————————  = 0 
               x2             

Step  3  :

When a fraction equals zero :

 3.1    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:

  (x+1)•(x-1)•(x2-7)
  —————————————————— • x2 = 0 • x2
          x2        

Now, on the left hand side, the  x²  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (x+1)  •  (x-1)  •  (x²-7)  = 0

Theory - Roots of a product :

 3.2    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 :

 3.3      Solve  :    x+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x = -1

Solving a Single Variable Equation :

 3.4      Solve  :    x-1 = 0 

 Add  1  to both sides of the equation : 
                      x = 1

Solving a Single Variable Equation :

 3.5      Solve  :    x²-7 = 0 

 Add  7  to both sides of the equation : 
                      x² = 7
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 7  

 The equation has two real solutions  
 These solutions are  x = ± √7 = ± 2.6458  
 

Supplement : Solving Quadratic Equation Directly

Solving    x4-8x2+7  = 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

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 4.1     Solve   x⁴-8x²+7 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that  w = x²  transforms the equation into :
 w²-8w+7 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   7.0000  or   1.0000

Now that we know the value(s) of  w , we can calculate  x  since  x  is  √ w  

Doing just this we discover that the solutions of
   x⁴-8x²+7 = 0
  are either : 
  x =√ 7.000 = 2.64575  or :
  x =√ 7.000 = -2.64575  or :
  x =√ 1.000 = 1.00000  or :
  x =√ 1.000 = -1.00000

Four solutions were found :

1.  x = ± √7 = ± 2.6458
2.  x = 1
3.  x = -1