Question

$(\alpha + \beta)(\delta + \gamma) = k - 30$

$\alpha + \beta + \gamma + \delta = 18$

$\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = -200$

$\alpha \beta = -32$, $\gamma \delta = 62$

Find $k$

$\alpha, \beta, \gamma, \delta$ are variables finite numbers.

Answer 1

86

Answer 2

Let

$S_1 = \alpha + \beta, \quad S_2 = \gamma + \delta$.

Then, from the second equation,

$S_1 + S_2 = 18 \quad \Rightarrow \quad S_2 = 18 - S_1$.

The triple product sum can be rewritten as:

$\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = \alpha\beta(\gamma+\delta) + \gamma\delta(\alpha+\beta) = (-32)S_2 + 62S_1$.

Given that this sum equals -200,

$-32(18-S_1) + 62S_1 = -200$.

Expanding,

$-576 + 32S_1 + 62S_1 = -200 \quad \Rightarrow \quad -576 + 94S_1 = -200$.

Solving for $S_1$,

$94S_1 = 376 \quad \Rightarrow \quad S_1 = 4$.

Then,

$S_2 = 18 - 4 = 14$.

We are also given,

$(\alpha + \beta)(\gamma + \delta) = k - 30$.

Substitute $S_1$ and $S_2$,

$4 \times 14 = k - 30 \quad \Rightarrow \quad 56 = k - 30$.

Thus,

$k = 86$.