Question

If $\alpha, \beta, \gamma, \sigma$ are the roots of the equation

$x^4 + 4x^3 - 6x^2 + 7x - 9 = 0$ then the value of

$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\sigma^2)$ is

Answer 1

13

Answer 2

The given polynomial equation is $P(x) = x^4 + 4x^3 - 6x^2 + 7x - 9 = 0$. Let its roots be $\alpha, \beta, \gamma, \sigma$.

Since the leading coefficient of the polynomial is 1, we can express $P(x)$ in terms of its roots as: $P(x) = (x-\alpha)(x-\beta)(x-\gamma)(x-\sigma)$.

We need to find the value of the product $(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\sigma^2)$. We can rewrite each term $(1+r^2)$ as $(r-i)(r+i)$, where $i = \sqrt{-1}$. So, the expression becomes:

$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\sigma^2) = (\alpha-i)(\alpha+i)(\beta-i)(\beta+i)(\gamma-i)(\gamma+i)(\sigma-i)(\sigma+i)$

Rearrange the terms by grouping those with $(r-i)$ and $(r+i)$:

$= [(\alpha-i)(\beta-i)(\gamma-i)(\sigma-i)] \times [(\alpha+i)(\beta+i)(\gamma+i)(\sigma+i)]$

Now, let's relate these products to the polynomial $P(x)$:

1. Consider $P(i)$:

$P(i) = (i-\alpha)(i-\beta)(i-\gamma)(i-\sigma)$.

We can factor out $(-1)$ from each term in the first bracket of our expression:

$(\alpha-i)(\beta-i)(\gamma-i)(\sigma-i) = (-1)^4 (i-\alpha)(i-\beta)(i-\gamma)(i-\sigma) = P(i)$.

2. Consider $P(-i)$:

$P(-i) = (-i-\alpha)(-i-\beta)(-i-\gamma)(-i-\sigma)$.

We can factor out $(-1)$ from each term in the second bracket of our expression:

$(\alpha+i)(\beta+i)(\gamma+i)(\sigma+i) = (-1)^4 (-\alpha-i)(-\beta-i)(-\gamma-i)(-\sigma-i) = P(-i)$.

Therefore, the value we need to find is $P(i) \times P(-i)$.

First, calculate $P(i)$:

$P(i) = i^4 + 4i^3 - 6i^2 + 7i - 9$

Using the properties of $i$: $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

$P(i) = (1) + 4(-i) - 6(-1) + 7i - 9$

$P(i) = 1 - 4i + 6 + 7i - 9$

$P(i) = (1+6-9) + (-4+7)i$

$P(i) = -2 + 3i$.

Next, calculate $P(-i)$:

$P(-i) = (-i)^4 + 4(-i)^3 - 6(-i)^2 + 7(-i) - 9$

$P(-i) = (1) + 4(i) - 6(-1) - 7i - 9$

$P(-i) = 1 + 4i + 6 - 7i - 9$

$P(-i) = (1+6-9) + (4-7)i$

$P(-i) = -2 - 3i$.

Finally, multiply $P(i)$ and $P(-i)$:

$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\sigma^2) = P(i) \times P(-i)$

$= (-2 + 3i)(-2 - 3i)$

This is in the form $(a+bi)(a-bi) = a^2 + b^2$.

$= (-2)^2 + (3)^2$

$= 4 + 9$

$= 13$.