# If 3+i is a root of the equation y=x^4-3x^3-36x^2+198x-280, find the sum of the squares of the real roots?

Sep 17, 2017

$65$

#### Explanation:

Note that if the zeros are $\alpha , \beta , \gamma , \delta$ then:

${x}^{4} - 3 {x}^{3} - 36 {x}^{2} + 198 x - 280$

$= \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) \left(x - \delta\right)$

$= {x}^{4} - \left(\alpha + \beta + \gamma + \delta\right) {x}^{3} + \left(\alpha \beta + \beta \gamma + \gamma \delta + \delta \alpha + \alpha \gamma + \beta \delta\right) {x}^{2} - \left(\alpha \beta \gamma + \beta \gamma \delta + \gamma \delta \alpha + \delta \alpha \beta\right) x - \alpha \beta \gamma \delta$

Equating coefficients, we find:

$\left\{\begin{matrix}\alpha + \beta + \gamma + \delta = 3 \\ \alpha \beta + \beta \gamma + \gamma \delta + \delta \alpha + \alpha \gamma + \beta \delta = - 36 \\ \alpha \beta \gamma + \beta \gamma \delta + \gamma \delta \alpha + \delta \alpha \beta = 198 \\ \alpha \beta \gamma \delta = 280\end{matrix}\right.$

Note that:

${\alpha}^{2} + {\beta}^{2} + {\gamma}^{2} + {\delta}^{2}$

$= {\left(\alpha + \beta + \gamma + \delta\right)}^{2} - 2 \left(\alpha \beta + \beta \gamma + \gamma \delta + \delta \alpha + \alpha \gamma + \beta \delta\right)$

$= {3}^{2} - 2 \left(- 36\right)$

$= 9 + 72$

$= 81$

Now since the coefficients of the quartic are all real, both $3 + i$ and $3 - i$ are zeros, with the sum of their squares being:

${\left(3 + i\right)}^{2} + {\left(3 - i\right)}^{2} = \left(9 + 6 i - 1\right) + \left(9 - 6 i - 1\right) = 16$

So the sum of the squares of the other two zeros is:

$81 - 16 = 65$

Note that by Descartes' Rule of Signs we can tell that the quartic has one negative zero and one or three positive zeros. Since we know that two zeros are non-real, that leaves one positive zero and one negative one.

Sep 17, 2017

$65$

#### Explanation:

Given:

${x}^{4} - 3 {x}^{3} - 36 {x}^{2} + 198 x - 280$

and one of the zeros:

$3 + i$

First note that all the coefficients of the given quartic are real. So any non-real zeros must occur in complex conjugate pairs.

So another zero must be:

$3 - i$

So we can find a quadratic factor of the quartic as:

$\left(x - \left(3 + i\right)\right) \left(x - \left(3 - i\right)\right) = \left(\left(x - 3\right) + i\right) \left(\left(x - 3\right) - i\right)$

$\textcolor{w h i t e}{\left(x - \left(3 + i\right)\right) \left(x - \left(3 - i\right)\right)} = {\left(x - 3\right)}^{2} - {i}^{2}$

$\textcolor{w h i t e}{\left(x - \left(3 + i\right)\right) \left(x - \left(3 - i\right)\right)} = {x}^{2} - 6 x + 9 + 1$

$\textcolor{w h i t e}{\left(x - \left(3 + i\right)\right) \left(x - \left(3 - i\right)\right)} = {x}^{2} - 6 x + 10$

Separating out this factor we find:

${x}^{4} - 3 {x}^{3} - 36 {x}^{2} + 198 x - 280 = \left({x}^{2} - 6 x + 10\right) \left({x}^{2} + 3 x - 28\right)$

To factor ${x}^{2} + 3 x - 28$ find a pair of factors of $28$ which differ by $3$.

The pair $7 , 4$ works.

So we find:

${x}^{2} + 3 x - 28 = \left(x + 7\right) \left(x - 4\right)$

So the real zeros are $- 7$ and $4$.

The sum of their squares is:

${\left(- 7\right)}^{2} + {4}^{2} = 49 + 16 = 65$