Solve equation in real number system

Question

Solve the equation in the real number system 

x^4-32x^2-144=0

Answer 1

x^4-32x^2 -144=0, 

 

  To factorize, we need to express -32 as sum of two numbers such that product of those two numbers is -144.

 

 -32 = -36 +4 , and -36 * 4 = -144

 

Given equation is: X^4 -36x^2 +4X^ -36*4 = 0

   (x^2 -36 ) ( x^2 + 4) = 0.  Now for any real x, x^2 >= 0, so x^2 + 4 >= 4

  The equation yields x^2 -36 = 0 => x = +6 or x = -6

 

  

Answer 2

This equation has 2 real roots and 2 imaginary roots. It asks for only the solution in the real number system.

There are 2 real answers, 6 and -6.   The Imaginary solutions are 2i and -2i where i is the square root of -1.

 

Solution: Let y = x^2   Then x^4 - 32 x^2 - 144 = 0 can be rewritten as y^2 -32y - 144 =0

 

Now y^2 -32y -144 can be factored into (y + 4) (y - 36).

Since y = x^2 this factorization becomes (x^2 + 4) (x^2 - 36)

 

Now, If x^2 -36 = 0 then x = 6 or -6 (These are the real roots of the equation.)

 

If x^2 + 4 = 0 then x = the square root of -4 which is 2i or -2i. (These are the imaginary roots)

 

If you need a further explanation, write back.

 

Answer 3

Jessica, you could use the foil method to solve for x in this equation. It looks like the factors for 144 in which their difference being 32 are 36 and 4.

 

x^4-32x^2-144=0

(x^2-36)(x^2+4)=0 Now, you could further factor the first binomial in which is a perfect difference in square.

(x-6)(x+6)(x^2+4)=0

 

This implies the real values for x are x=6 and x=-6. The rest are two imaginary numbers in which x=±2i