What's the most efficient way to calculate the zeros of $f(x) = x^4-x^2-6$?

Question

How do you calculate the zeros of $f(x) = x^4-x^2-6$ in best way? Here are my attempts:

Factorize, but the problem is the $6$:

$$0=x^4-x^2-6 \iff 6 = x^2(x^2-1)$$

This doesn't lead to any good solution...

Here is another attempt, but I don't know if this is allowed:

$$0= x^4-x^2-6 \iff 0 = \sqrt{x^4} - \sqrt{x^2} - \sqrt{6}$$

My friend say this no work because I cheat. I Need take minus signs under square root too. Is this tru? If not I can continue this with quadratic equation formula and get the zeroes.

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Please halp us this is no homework he make this Task because he is curious how is solved and me curious also now. Because our teacher only ask easy Task but we want other Task also because his lesson is always easy but exam is very hard when you compared...

And also pls halp with maths Topic I'm not sure which is corrects.

Answer 1

Once you got to $x^2 (x^2-1) = 6$ (in plain English, "a number times (itself - $1$) $= 6$), you could have thought that $6 = 3 \cdot 2$ , and conclude in one go, finding the two real roots $\pm \sqrt{3}$.

Answer 2

after my hint above we have to solve $$t^2-t-6=0$$ this is $$t_{1,2}=\frac{1}{2}\pm\sqrt{\frac{1}{4}+\frac{24}{4}}$$

Answer 3

SUbstitute $x^2=u$ so you get a quadratic equation

$u^2-u-6=0$

then you solve and you get two solutions, one is negative and when you substitute back you have $x^2=negative$ which leads to complex roots

Hope this helps

Answer 4

Solve the equation $(x^2)^2-x^2-6=0$. You will get$$x^2=\frac{1\pm\sqrt{25}}2\iff x^2=3\vee x^2=-2.$$But $x^2$ cannot be $-2$. Therefore, $x^2=3$, and this means that the solutions are $\pm\sqrt3$.