Vrftsjtryk

So the question is this:

Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:

begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}

I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:

Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.

begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.

Working out the values of $U_2-U_1$ and $U_3-U_2$:

begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}

That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?

Let's see:

$2x^4−68x^2−1468=x^4−8x^2−2007$

$y = x^2$

$2y^2-68y-1468=y^2-8y-2007$

$y^2-60y-539=0$

$(y-30)^2-539=900$

$(y-30)^2=1439$

$y-30=pm 1439^{1/2}$

$y = 30pm 1439^{1/2}$

$x = 1469^{1/2}$

$x = 1409^{1/2}$

Pressed enter accidentally up there and it looks weird as a comment.

Since $ 2007-1468=539$ you have

$(y-30)^2color{red}+539=900quad |-539$

$(y-30)^2=361quad | sqrt{()} $

$y-30=pm 19$

$
y_1=11, y_2=49$

Re-subtitution $sqrt{y}=pm x$

$x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$