x^4 -37x^2 = -36
I would like to know how to solve it and the answer. Don't just give the answer I need to understand what is going on.
-Thanks for any help
x^4 -37x^2 = -36 (equation 1)
This problem is easy and can be solved in many different ways :
1-Smart way requires some thought process but much faster and exceptionally works here (I'll explain the thought process step by step): you'd notice immediately that 1 and -1 are evident solutions to the equation (always try this when you are solving polynomials if 0, 1, -1 if they are obvious solutions or not)
So this means you can factor by those solution and equation1 becomes (x-1)(x+1)(ax^2+bx+c)=0 (equation 2) btw (x-1)(x+1)= x^2-1 (remarkable identity, I can say easily that a=1 since no factor in front of x^4 (identification factor by factor in a polynomial equation aka if ax^n+bx^(n-1)....=ux^n+ix^(n-1)... for every x from R <=> a=u, b=i,.... ), c is also easy if -1*1*c=36 hence c=-36 all is left is b, well since in eq1 there is x^3 or x by default I can say b=0
So we have (x-1)(x+1)(x^2-36) =0 and we easily notice here that we have another remarkable identity x^2-36= (x-6)(x+6)
So at the end we have (x-1)(x+1)(x+6)(x-6)=0 hence -6 -1 1 6 are the solution of the equation
tl,dr :
x^4 -37x^2 +36 = 0 has -1 and 1 as obvious solutions so you can factorize by (x+1)(x-1) so the equation becomes (x+1)(x-1)(ax^2+bx+c)=0 a, b and c are easily identifiable (using the identification factor by factor method), a=1 b=0 c=-36 so we have (x+1)(x-1)(x^2-36)=0
x^2-36 is a remarkable identity (x-6)(x+6)=x^2-36 so the equation becomes (x-1)(x+1)(x+6)(x-6)=0 and you've got your solutions
2-More traditional way which is explained by Foxpup so I refer you to his reply which is by using a variable change by putting x^2=z and solving the problem as a second degree polynomial
