Thread: Two exercises - polynomial equations and geometric and arithmetic sequences

1. Two exercises - polynomial equations and geometric and arithmetic sequences

Hey, my neighbour again send me one exercise.

1. In the equation $x^{3} - 7x^{2} - 21x + a = 0$ you must find solutions whose are in geometric sequence. For which $a$?

How can I find this $a$?

2. In the equation $x^{4} -(a+3)x^2 + (a + 2) = 0$ must find this $a$ that solutions are in arithmetic sequence.

How can I find this $a$?

2. Re: Two exercises - polynomial equations and geometric and arithmetic sequences

(1) A cubic equation can have up to 3 distinct solutions. If they are in geometric progression, we can write them as u, ur, and ur^2. If those are the solutions to a cubic equation with this equation, you must have (x- u)(x- ur)(x- ur^2)= x^3- 7x^2- 21x+ a. Multiply out on right and compare the coefficients of x^2 and x on left and right to get two equations for u and r. Once you have those it is easy to find a.

(2) Pretty much the same thing. A quartic equation can have up to 4 distinct solutions. If the solutions are in arithmetic progression, we can write them as u, u+ r, u+ 2r, and u+ 3r. So we must have (x- u)(x- u- r)(x- u- 2r)(u+ 3r)= x^4- (a+ 3)r^2+ (a+ 2).

3. Re: Two exercises - polynomial equations and geometric and arithmetic sequences

Hello, lebdim!

1. In the equation $x^3 - 7x^2 - 21x + b = 0$, the solutions are in geometric sequence.
. . For which $b$?

The roots are: . $a,\,ar,\,ar^2.$

Using Vieta's formulas, we have:

. . $\begin{Bmatrix}a+ar+ar^2 \:=\:7 & \Rightarrow & a(1+r+r^2) \:=\:7 & [1] \\ a^2r + a^2r^2 + a^2r^3 \:=\:-21 & \Rightarrow& a^2r(1+r+r^2) \:=\:-21 & [2] \\ a^3r^3 \:=\:-b & \Rightarrow & (ar)^3 \:=\:-b & [3] \end{Bmatrix}$

Divide [2] by [1]: . $\frac{a^2r(1+r+r^2)}{a(1+r+r^2)} \:=\:\frac{-21}{7} \quad\Rightarrow\quad ar \:=\:-3$

Cube both sides: . $(ar)^3 \:=\:(-3)^3 \:=\:-27$

Substitute into [3]: . $-b \:=\:-27 \quad\Rightarrow\quad \boxed{b \:=\:27}$