solve for $x$: $(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$

Question

There is an equation that I think it is complicated ,a little! $$(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$$ Actually we must solve for $x$ here. I want you to hint me how can I simplify the equation and solve it.

Answer 1

It's not as hard as it first looks. Note that all of the summands are non-negative and their sum is zero, which can only happen if they are all zero.

Answer 2

Hint:

Observe that for real $x$

each of the summands must be $=0$

Now $x^2+x-6=(x+3)(x-2)$

$$x^4-13x^2+36=(x^2-4)(x^2-9)$$

$$x^3-7x+6=x^3-2^3-7(x-2)$$ or try divide $x^3-7x+6$ by $x-2$

Answer 3

Hint:

The first expression is biquadratic and

$$x^4-13x^2+36=(x^2-4)(x^2-25).$$

The second is quadratic,

$$x^2+x-6=(x-2)(x+3).$$

The third is cubic and by inspection $x=1$ is a root. So

$$x^3-7x+6=(x-1)(x^2+x-6)=(x-1)(x-2)(x+3).$$

Now you need to find the common roots.