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Solution - Equations reducible to quadratic form

x=3
x=3
x=3
x=-3
x=0.00002.2361i
x=0.0000-2.2361i
x=0.0000+2.2361i
x=0.0000+2.2361i

Step by Step Solution

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((x4) -  22x2) -  45  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  x4-4x2-45 

The first term is,  x4  its coefficient is  1 .
The middle term is,  -4x2  its coefficient is  -4 .
The last term, "the constant", is  -45 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -45 = -45 

Step-2 : Find two factors of  -45  whose sum equals the coefficient of the middle term, which is   -4 .

     -45   +   1   =   -44
     -15   +   3   =   -12
     -9   +   5   =   -4   That's it


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  5 
                     x4 - 9x2 + 5x2 - 45

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x2 • (x2-9)
              Add up the last 2 terms, pulling out common factors :
                    5 • (x2-9)
Step-5 : Add up the four terms of step 4 :
                    (x2+5)  •  (x2-9)
             Which is the desired factorization

Polynomial Roots Calculator :

 2.2    Find roots (zeroes) of :       F(x) = x2+5
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  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  5.

 
The factor(s) are:

of the Leading Coefficient :  1
 
of the Trailing Constant :  1 ,5

 
Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor
     -1     1      -1.00      6.00   
     -5     1      -5.00      30.00   
     1     1      1.00      6.00   
     5     1      5.00      30.00   


Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Squares :

 2.3      Factoring:  x2-9 

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 :  x2  is the square of  x1 

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

Equation at the end of step  2  :

  (x2 + 5) • (x + 3) • (x - 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  :    x2+5 = 0 

 
Subtract  5  from both sides of the equation : 
 
                     x2 = -5
 
 
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
 
                     x  =  ± √ -5  

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

Accordingly,  √ -5  =
                    √ -1• 5   =
                    √ -1 •√  5   =
                    i •  √ 5

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

                      x=  0.0000 + 2.2361
                      x=  0.0000 - 2.2361

Solving a Single Variable Equation :

 3.3      Solve  :    x+3 = 0 

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

Solving a Single Variable Equation :

 3.4      Solve  :    x-3 = 0 

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

Supplement : Solving Quadratic Equation Directly

Solving    x4-4x2-45  = 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   x4-4x2-45 = 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 :
 w2-4w-45 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   9.0000  or  -5.0000

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
   x4-4x2-45 = 0
  are either : 
  x =√ 9.000 = 3.00000  or :
  x =√ 9.000 = -3.00000  or :
  x =√-5.000 = 0.0 + 2.23607 i  or :
  x =√-5.000 = 0.0 - 2.23607 i

Four solutions were found :

  1.  x = 3
  2.  x = -3
  3.   x=  0.0000 - 2.2361
  4.   x=  0.0000 + 2.2361

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