Mathematico | Mathematico

The equation

{(x^{2} + 3 x + 2)}^{2} - 8 (x^{2} + 3 x + 1) + 4 = 0

has $n$ solutions, and the sum of these solutions is $S$ .

Find $S^{n}$ .

Hint

This is almost a quadratic in disguise... we just need to make a minor adjustment

Hint

Try this:

{(x^{2} + 3 x + 2)}^{2} - 8 ([x^{2} + 3 x + 2] - 1) + 4 = 0

Hint

So the equation becomes:

{(x^{2} + 3 x + 2)}^{2} - 8 (x^{2} + 3 x + 2) + 12 = 0

Now let

u = x^{2} + 3 x + 2

and keep solving!

Solution

We make an adjustment by sticking a $+ 1$ in the second bracket. After expanding out, this means we've put in a $- 8$ and so we need to $+ 8$ outside the bracket to balance.

\begin{aligned} {(x^{2} + 3 x + 2)}^{2} - 8 (x^{2} + 3 x + 1) + 4 & = 0 \\ {(x^{2} + 3 x + 2)}^{2} - 8 (x^{2} + 3 x + 2) + 8 + 4 & = 0 \\ {(x^{2} + 3 x + 2)}^{2} - 8 (x^{2} + 3 x + 2) + 12 & = 0 \end{aligned}

Now let $u = x^{2} + 3 x + 2$

\begin{aligned} u^{2} - 8 u + 12 & = 0 \\ (u - 6) (u - 2) & = 0 \end{aligned}

So we get

\begin{aligned} u & = 6 \\ x^{2} + 3 x + 2 & = 6 \\ x^{2} + 3 x - 4 & = 0 \\ (x + 4) (x - 1) & = 0 \\ x & \in {- 4, 1} \end{aligned}

and

\begin{aligned} u & = 2 \\ x^{2} + 3 x + 2 & = 2 \\ x^{2} + 3 x & = 0 \\ x (x + 3) & = 0 \\ x & \in {- 3, 0} \end{aligned}

So the full solution set is

x \in {- 4, - 3, 0, 1}