Relationships between roots and co-efficients 0: (1 Viewer)

Kiriii cx · Jul 23, 2018

Helppppppp….
I stumbled across these questions :c

14. If the roots of the equation x^3 + 3x^2 - 2x + 1 = 0 are α,β ,γ, find the value of:

(a) α^2 (β + γ) + β ^2 (γ + α) + γ^2 (α + β)

(b) (α^2)(β ^2) + (β ^2) (γ^2) + (γ^2)(α^2)

Answers are: (a) (9) (b) (-2)

16. Solve the equation 6x^4 - 11x^3 - 26x^2 + 22x + 24 = 0 given that the product of two
of its roots is equal to the product of the other two roots.

Answer is: 4/3 , -3/2 , (1+root3) , (1- root3)

Please show full solution, really appreciate any help ^_^

fan96 · Jul 23, 2018

Note that

$\begin{aligned} \alpha + \beta + \gamma &= -3 \\ \alpha \beta +\alpha \gamma + \beta \gamma &=-2\\ \alpha \beta \gamma &=-1 \end{aligned}$

14a)

Let the expression in a) be denoted as

$\sum (\alpha_i)^2 (\alpha_j)$ .

Now,

$\begin{aligned}(\alpha + \beta + \gamma) (\alpha \beta +\alpha \gamma + \beta \gamma) &= 3\alpha\beta\gamma + \sum (\alpha_i)^2 (\alpha_j) \\ 6 &= -3 + \sum (\alpha_i)^2 (\alpha_j) \\ \sum (\alpha_i)^2 (\alpha_j) &= 9\end{aligned}$

14b)

Let the expression in b) be denoted as

$\sum_{i<j} (\alpha_i \alpha_j)^2$ .

Now,

$\begin{aligned} (\alpha \beta +\alpha \gamma + \beta \gamma)^2 &= \sum_{i<j} (\alpha_i \alpha_j)^2 + 2\alpha\beta\gamma(\alpha+\beta+\gamma) \\ \sum_{i<j} (\alpha_i \alpha_j)^2 &= -2\end{aligned}$

16)

Define some equations, which can be derived using sum of roots and the given condition $\alpha\beta = \gamma\delta$ :

$(1): (\alpha\beta)^2 = 4$

$(2):\delta = \frac{\alpha\beta}{\gamma}$

$(3): \alpha+\beta+\gamma= \frac{11}{6} - \frac{\alpha\beta}{\gamma}$

$(4): \delta+\gamma=\frac{11}{6} - (\alpha+\beta)$

Sum of roots three at a time:

$\alpha\beta\gamma + \alpha\gamma\delta + \beta\gamma\delta + \alpha\beta\delta = -\frac{11}{3}$

Substituting equation (2) and simplifying ends up with

$\alpha\beta\left(\alpha+\beta+\gamma\right) + \frac 4 \gamma= -\frac{11}{3}$

Substituting (3) gives us

$\alpha\beta\left(\frac{11}{6} - \frac{\alpha\beta}{\gamma}\right) + \frac 4 \gamma= -\frac{11}{3}$

and simplifying again using (1) ends up with

$(5): \alpha\beta = -2 = \delta\gamma$ .

Substituting (5) into sum of roots two at a time and factorising gives:

$-4 + (\alpha + \beta)(\gamma + \delta) = -\frac{13}{3}$

Substitute (4) and simplify:

$(\alpha + \beta)\left(\frac{11}{6} - (\alpha + \beta)\right) = -\frac{1}{3}$

This is a quadratic in $\alpha + \beta$ . If we solve and take the positive root we obtain:

$(6): \alpha + \beta = 2$

Solve (5) and (6) to get the first two roots:

$1 + \sqrt 3$ and $1 - \sqrt 3$

And from there it is easy to obtain the last two roots.

Kiriii cx · Jul 24, 2018

Wow, thanks so much, I get the second question now.
However for the first question, the process you used... I didn't fully understand how you did it.
Why did you use the sigma notation for substituting the expression itself?

Thanks again for all this help

fan96 · Jul 24, 2018

Kiriii cx said:
Why did you use the sigma notation for substituting the expression itself?

Because it's a huuuuge pain to write out the entire thing in TeX and using the sigma notation is just faster.

Kiriii cx · Jul 27, 2018

Aye, so I can choose not to use it, because somehow the double sigma scares me XD

Appreciate your help

Relationships between roots and co-efficients 0: (1 Viewer)

Kiriii cx

New Member

fan96

617 pages

Kiriii cx

New Member

fan96

617 pages

Kiriii cx

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)