0
$\begingroup$

I have had trouble with this question:

"Solve the equation $8x^3 - 38x^2 + 57x -27 = 0$" if the roots are in geometric progression.

Any help would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ So what did you learn from math.stackexchange.com/questions/392309/… ? $\endgroup$ – Isomorphism May 18 '13 at 11:05
  • $\begingroup$ I got to the point where I divided 1+r+r^2, but when I try to form a new quadratic equation - I end up with an imaginary number. I'm only asking the question to see if I'm missing anything important. $\endgroup$ – missiledragon May 18 '13 at 11:06
  • $\begingroup$ Did you get the roots as 1, 9/4 and 3/2? $\endgroup$ – Isomorphism May 18 '13 at 11:20
  • $\begingroup$ Yes, I got them - but I got 3/2 as well as 2/3 in my quadratic equation. How do I choose? $\endgroup$ – missiledragon May 18 '13 at 12:12
  • $\begingroup$ Please help me, I'm stuck. $\endgroup$ – missiledragon May 18 '13 at 12:23
1
$\begingroup$

Let the roots be $a, a\cdot r, a\cdot r^2$

So using Vieta's formula $a+ a\cdot r+ a\cdot r^2=\frac{38}8\implies a(1+r+r^2)=\frac{19}4$

and $a( a\cdot r)+ a\cdot r(a\cdot r^2)+a\cdot r^2(a)=\frac{57}8\implies a^2r(1+r^2+r)=\frac{57}8$

Divide to get $ar=\frac{\frac{57}8}{\frac{19}4}=\frac32$ as $a(1+r+r^2)\ne0$

Put $a=\frac3{2r}$ in the first equation

$\endgroup$
  • $\begingroup$ Thanks, I know you already answered a similar question - but for some reason I ended with weird numbers. Thanks for clearing it up! $\endgroup$ – missiledragon May 18 '13 at 11:09
  • $\begingroup$ Now, what's wrong with this method, that caused the down-vote? $\endgroup$ – lab bhattacharjee May 18 '13 at 11:14
  • $\begingroup$ I didn't downvote, I'll upvote now. $\endgroup$ – missiledragon May 18 '13 at 11:18
  • $\begingroup$ The question/request is to the down-voter to disclose the mistake that eluded me $\endgroup$ – lab bhattacharjee May 18 '13 at 11:20
  • $\begingroup$ I got that r=3/2,2/3 in a quadratic equation. How do I tell which is correct? $\endgroup$ – missiledragon May 18 '13 at 11:46
1
$\begingroup$

Since the roots are in geometric progression, you have $x_2^2=x_1x_3$. From Theorem of Viet $x_1x_2x_3=27$. Hence $x_2^3=27$ etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.