Given $x^2 - 3x +6=0$ find $x^4$ in terms of $x$.

Question

Could someone please explain what this question is asking? I might be able to solve it by myself by what does it mean by $x^4$? I interpret it as provide a value of $x^4$ given this quadratic so quadratic formula then raise both sides to power $4$ ? There are no real solutions to this quadratic anyway as well so I am confused.

Answer 1

you can start from here $$x^2=3x-6$$ then multiply by $x$ $$x^3=3x^2-6x$$ now put $x^2=3x-6$ it means $$x^3=3(3x+6)-6x=9x+18-6x$$ now you have $x^3$ in terms of x.

can you take over?

Answer 2

I interpret it as

$x^2 -3x + 6 = 0$ so $x^2 = 3x -6$ so $x^4 = ????$ what?

$x^4 = (x^2)^2 = (3x-6)^2 = 9x^2 -36x + 36$

But we aren't done. Can we get $x^4 = ax + b$ for some $a$ and $b$?

$x^4 = 9x^2 -36x + 36 = 9(3x-6)-36x + 36 = 27x - 54 -36x + 36=-9x -18$

Answer 3

Your task is to find an equation of the form $$x^4=f(x)$$ where $f$ has the least possible degree, given the relation $x^2-3x+6=0$. Well, we can rearrange this to $x^2=3x-6$, then square both sides to get $x^4=(3x-6)^2=9x^2-36x+36$. Substituting $x^2=3x-6$ again into this gives the final answer as $9(3x-6)-36x+36=-9x-18$.

Answer 4

$x^2 = 3(x-2) \Rightarrow x^4 = 9(x-2)^2$

$= 9(x^2 - 4x + 4) = 9[3(x-2) - 4x + 4]$

$= 9[3x - 6 - 4x + 4] = 9[-x - 2].$