Topic: Help me with this simple math! (Read 608 times)

Thanks everyone I got it down now!

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

Tell that to Satoshi Nakamoto.

this is simple math ??  no wonder i never was good at maths ,, i hate maths ,, sorry cant help you ,, you are on your own mate ,,

Pay attention. Biquadratic equations can have up to four solutions. This particular one does, in fact, have four solutions. An answer which gives only one solution is therefore wrong. Now quit giving wrong answers and go back to school.

If your maths is as bad as me, You can use the plug in method and just try an error. Most of the time it would save you lots of time... So answer is 1

Wrong. Remember, biquadratic equations may have as many as four solutions.

Since it's a biquadratic equation, we can make it easier by letting z be x^2, so we've got a simple quadratic equation. Simply find the roots of the quadratic equation 1z^2-37z=-36:
  Add 36 to both sides:
  1z^2-37z+36 = 0
  Find the discriminant d = 37^2-4*1*36 = 1225
  Then solve for z: z = (-(-37)+d^(1/2))/(2*1)
  z+ = (37+35)/2  z- = (37-35)/2
  z+ = 36  z- = 1
Since z = x^2, the solutions to the original biquadratic equation are simply the square roots of z:
x++ = 6
x+- = -6
x-+ = 1
x-- = -1

So the four solutions are 6, -6, 1, and -1.

The answer is 1 

Sorry, I`m clueless on what that is?

Use a calculator?

That aint no simple math.

And you wont be using that anyways in the real world lol.

Here you go: https://www.khanacademy.org/math/algebra2/polynomial_and_rational/factoring-higher-deg-polynomials/v/factoring-special-products-2

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