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 :

                     w^4-5*w^2-(36)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((w4) -  5w2) -  36  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  w⁴-5w²-36 

The first term is,  w⁴  its coefficient is  1 .
The middle term is,  -5w²  its coefficient is  -5 .
The last term, "the constant", is  -36 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -36 = -36 

Step-2 : Find two factors of  -36  whose sum equals the coefficient of the middle term, which is   -5 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  4 
                     w⁴ - 9w² + 4w² - 36

Step-4 : Add up the first 2 terms, pulling out like factors :
                    w² • (w²-9)
              Add up the last 2 terms, pulling out common factors :
                    4 • (w²-9)
Step-5 : Add up the four terms of step 4 :
                    (w²+4)  •  (w²-9)
             Which is the desired factorization

Polynomial Roots Calculator :

 2.2    Find roots (zeroes) of :       F(w) = w²+4
Polynomial Roots Calculator is a set of methods aimed at finding values of  w  for which   F(w)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  w  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  4.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,4

 Let us test ....

Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Squares :

 2.3      Factoring:  w²-9 

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 :  w²  is the square of  w¹ 

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

Equation at the end of step  2  :

  (w2 + 4) • (w + 3) • (w - 3)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    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 :

 3.2      Solve  :    w²+4 = 0 

 Subtract  4  from both sides of the equation : 
                      w² = -4
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      w  =  ± √ -4  

 In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 

Accordingly,  √ -4  =
                    √ -1• 4   =
                    √ -1 •√  4   =
                    i •  √ 4

Can  √ 4 be simplified ?

Yes!   The prime factorization of  4   is
   2•2 
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).

√ 4   =  √ 2•2   =
                ±  2 • √ 1   =
                ±  2

The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      w=  0.0000 + 2.0000 i 
                      w=  0.0000 - 2.0000 i 

Solving a Single Variable Equation :

 3.3      Solve  :    w+3 = 0 

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

Solving a Single Variable Equation :

 3.4      Solve  :    w-3 = 0 

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

Supplement : Solving Quadratic Equation Directly

Solving    w4-5w2-36  = 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   w⁴-5w²-36 = 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  v , such that  v = w²  transforms the equation into :
 v²-5v-36 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   9.0000  or  -4.0000

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

Doing just this we discover that the solutions of
   w⁴-5w²-36 = 0
  are either : 
  w =√ 9.000 = 3.00000  or :
  w =√ 9.000 = -3.00000  or :
  w =√-4.000 = 0.0 + 2.00000 i  or :
  w =√-4.000 = 0.0 - 2.00000 i

Four solutions were found :

1.  w = 3
2.  w = -3
3.   w=  0.0000 - 2.0000 i 
   
4.   w=  0.0000 + 2.0000 i