ambiguity in solving algebraic equations.

Given $$x+\frac{1}{x}=2 \tag{1}$$
$$x^3+\frac{1}{x^4}=1 \tag{2}$$

Find the value of $$x^4+\frac{1}{x^3}$$

If equation $$\bf{2}$$ is multiplied by $$\bf x$$ we get $$x^4+\frac{1}{x^3}=x$$ now, we just need to find the value of $$x$$ which from $$\bf{1}$$ is equal to $$1$$. But the answer to this question is $$\bf3$$. Which can be get by solving differently. Why my method is wrong is it because I am introducing new root into the equation.

Another approach Given, $$(x + 1/x) = 2$$

Squaring both sides,

$$x^2 + 1/x^2 + 2 = 4$$

$$x^2 + 1/x^2 = 2$$

Let,

$$A = x^3 + 1/x^4$$

$$B = x^4 + 1/x^3$$

$$A + B = x^3 + 1/x^4 + x^4 + 1/x^3$$

$$A + B = x^3 + 1/x^3 + x^4 + 1/x^4$$

$$A + B = (x + 1/x)(x^2 + 1/x^2 – 1) + (x^2)2 + (1/x^2)2 + 2 – 2$$

$$A + B = 2(2 – 1) + (x^2 + 1/x^2)2 – 2$$

$$A + B = 2 + (2)2 – 2$$

$$A + B = 4$$

Given that $$A = 1$$

Then, $$B = 4 – 1 = 3$$

$$x^4 + 1/x^3 = 3$$

• If $x+\frac1x=2$ then $x=1$. Are you sure you have the question right? – Lord Shark the Unknown Jul 22 at 17:10

The system is inconsistent because the first equation gives $$x=1$$ which does not satisfy the second equation.
If $$x=3$$ and $$x$$ is solution to $$(1)$$, we have $$x + \frac{1}{x} = 2 \iff 3 + \frac 1 3 = 2$$ which is obviously wrong. There must be some mistake in the solution.