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jlacosta   0
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0 replies
1 viewing
jlacosta
Mar 2, 2025
0 replies
a/b + b/a never integer ?
MTA_2024   3
N 2 hours ago by ohiorizzler1434
Let $a$ and $b$ be 2 distinct positive integers.
Can $\frac a b +\frac b a $ be in an integer. Prove why ?
3 replies
MTA_2024
Yesterday at 3:08 PM
ohiorizzler1434
2 hours ago
2014 Community AIME / Marathon ... Algebra Medium #1 quartic
parmenides51   5
N 5 hours ago by CubeAlgo15
Let there be a quartic function $f(x)$ with maximums $(4,5)$ and $(5,5)$. If $f(0) = -195$, and $f(10)$ can be expressed as $-n$ where $n$ is a positive integer, find $n$.

proposed by joshualee2000
5 replies
parmenides51
Jan 21, 2024
CubeAlgo15
5 hours ago
Inequalities
sqing   4
N Yesterday at 5:52 PM by DAVROS
Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 3  $. Prove that
$$ a+b+a^2+b^2 \geq 28-6\sqrt{21}$$Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 4  $. Prove that
$$ a+b+a^2+b^2 \geq 10-4\sqrt 5$$Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 5  $. Prove that
$$ a+b+a^2+b^2 \geq \frac {2(26-5\sqrt{13})}{9} $$Let $ a, b\geq 0 $ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq 3+ \sqrt {5}   $. Prove that
$$ a+b+a^2+b^2 \geq 2$$
4 replies
sqing
Mar 14, 2025
DAVROS
Yesterday at 5:52 PM
interesting problem
sausagebun   1
N Yesterday at 4:05 PM by mathprodigy2011
Six points, labeled A, B, C, D, E, and F, are positioned consecutively on a straight line. Let G be a point not located on this line. The following distances are given: AC = 26, BD = 22, CE = 31, DF = 33, AF = 73, CG = 40, and DG = 30. Determine the area of triangle BGE.
I brute forced this with trig, was wondering if theres a more elegant way of doing this
1 reply
sausagebun
Yesterday at 3:21 PM
mathprodigy2011
Yesterday at 4:05 PM
No more topics!
The third root
Silverfalcon   32
N Jun 20, 2020 by gabrupro
The roots of the equation $x^2+4x-5 = 0$ are also the roots of the equation $2x^3+9x^2-6x-5 = 0$. What is the third root of the second equation?
32 replies
Silverfalcon
Aug 15, 2005
gabrupro
Jun 20, 2020
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Silverfalcon
5006 posts
#1 • 2 Y
Y by Adventure10, Mango247
The roots of the equation $x^2+4x-5 = 0$ are also the roots of the equation $2x^3+9x^2-6x-5 = 0$. What is the third root of the second equation?
This post has been edited 1 time. Last edited by Silverfalcon, Jan 26, 2006, 2:13 PM
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236factorial
4555 posts
#2 • 2 Y
Y by Adventure10, Mango247
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h78reg
384 posts
#3 • 2 Y
Y by Adventure10, Mango247
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Silverfalcon
5006 posts
#4 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Great.

But here's another thought. Did we need to find the roots of the first equation?

What if the roots were something like $4+3i$ and $4-3i$? That wouldn't be a neat numbers to use in a synthetic division. Is there anyway we can do this problem WITHOUT using synthetic division?
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frt
1294 posts
#5 • 2 Y
Y by Adventure10, Mango247
We could simply do a long division ;)

$\frac{2x^3+9x^2-6x-5}{x^2+4x-5}=2x+1$
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Silverfalcon
5006 posts
#6 • 2 Y
Y by Adventure10, Mango247
:) Yes.

Although it doesn't hurt you to find the roots, frt's method is quicker and what the CMO solution said (and how I did.. I actually did frt's method and used synthetic division to check my answer).
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h78reg
384 posts
#7 • 2 Y
Y by Adventure10, Mango247
i think that's similar to what 236factorial did.
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236factorial
4555 posts
#8 • 2 Y
Y by Adventure10, Mango247
strange way
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Silverfalcon
5006 posts
#9 • 2 Y
Y by Adventure10, Mango247
That's not a strange way.

It is pretty clever actually. Noticing the first equation is MONIC (leading coefficent is 1) and the second equation has coefficent of 2 helps you to know that it's in form of $(2x+d)$. From here, like you did, it's not that hard to figure it out.

Good job! :)
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frt
1294 posts
#10 • 2 Y
Y by Adventure10, Mango247
Yeah, I think that's actually faster than doing a long division or synthetic division.
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ajai
191 posts
#11 • 2 Y
Y by Adventure10, Mango247
for synthetic division I thought that you can only divide by (x-a), how do you divide by
$x^2+4x+5$?
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236factorial
4555 posts
#12 • 2 Y
Y by Adventure10, Mango247
ajai wrote:
for synthetic division I thought that you can only divide by (x-a), how do you divide by
$x^2+4x+5$?

Do you mean $x^2+4x-5$?
Factoring, that makes (x+5) and (x-1). Now you can use synthetic division.
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Silverfalcon
5006 posts
#13 • 2 Y
Y by Adventure10, Mango247
I used Long Division.
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Iversonfan2005
1697 posts
#14 • 2 Y
Y by Adventure10, Mango247
Silverfalcon wrote:
The roots of the equation $x^2+4x-5 = 0$ are also the roots of the equation $2x^3+9x^2-6x-5 = 0$. What is the third root of the second equation?

Please post full solution to the problem. If you consider this problem easy, leave to other people who can benefit from trying this.

$x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$.
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10000th User
3049 posts
#15 • 2 Y
Y by Adventure10, Mango247
Let me add that this question can be solved by another method too. Solution by Viete formulas

1) Sum of roots of quadratic is -4. Sum of roots of cubic is $-4+r=-\frac92$. Third root is $r=-\frac12$

2) Product of roots of quadratic is -5. Product of roots of cubic is $-5r=\frac52$. Third root is $r=-\frac12$
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AntonioMainenti
416 posts
#16 • 2 Y
Y by Adventure10, Mango247
Iversonfan2005 wrote:
$x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$.
I hate to say this over and over, but this is a misuse of the $\Longrightarrow$ symbol. Statements imply other statements. An expression on its own cannot imply anything.
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Iversonfan2005
1697 posts
#17 • 2 Y
Y by Adventure10, Mango247
AntonioMainenti wrote:
Iversonfan2005 wrote:
$x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$.
I hate to say this over and over, but this is a misuse of the $\Longrightarrow$ symbol. Statements imply other statements. An expression on its own cannot imply anything.

Does it really matter? ;) :|
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10000th User
3049 posts
#18 • 2 Y
Y by Adventure10, Mango247
Iversonfan2005 wrote:
AntonioMainenti wrote:
Iversonfan2005 wrote:
$x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$.
I hate to say this over and over, but this is a misuse of the $\Longrightarrow$ symbol. Statements imply other statements. An expression on its own cannot imply anything.

Does it really matter? ;) :|
It doesn't matter a lot if used in forums, but I think it matters elsewhere :D
OK here is one suggestion to use it better:

$x^2+4x-5=(x-1)(x+5)=0\Longrightarrow x=1, -5$

$(x-1)(2x^2+11x+5)=0\Longrightarrow x-1=0\text{ or }2x^2+11x+5=0\Longrightarrow x=1\text{ or }(2x+1)(x+5)=0\Longrightarrow x=1,-5,\boxed{-\frac{1}{2}}$
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AntonioMainenti
416 posts
#19 • 2 Y
Y by Adventure10, Mango247
Iversonfan2005 wrote:
Does it really matter? ;) :|
Well, you're using an implication symbol where you really mean $=$. What if someone used $+$ where they meant $-$, or $\in$ where they meant $\le$. It's not any different to me at least. Anyway, I'm sure your math professors won't like it either, as 10000th User hinted at.
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#H34N1
4235 posts
#20 • 2 Y
Y by Adventure10, Mango247
what is a third root. :?
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eryaman
1130 posts
#21 • 1 Y
Y by Adventure10
anirudh wrote:
what is a third root. :?
Cubic equations have $3$ roots. In the question it gives you the first two roots (same as the quadratic equation's) and asks for the remaining (3rd) root.
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#H34N1
4235 posts
#22 • 2 Y
Y by Adventure10, Mango247
What is synthetic divisioin?
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Iversonfan2005
1697 posts
#23 • 2 Y
Y by Adventure10, Mango247
anirudh wrote:
What is synthetic divisioin?

It's a way of factoring by using trial and error using the last term over the first term to find the roots.
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#H34N1
4235 posts
#24 • 2 Y
Y by Adventure10, Mango247
but i don't like trial and error.
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Iversonfan2005
1697 posts
#25 • 2 Y
Y by Adventure10, Mango247
anirudh wrote:
but i don't like trial and error.

You must like spamming though.. ;)

Go read about synthetic division, it's very useful.
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236factorial
4555 posts
#26 • 1 Y
Y by Adventure10
Iversonfan2005 wrote:
anirudh wrote:
but i don't like trial and error.

You must like spamming though.. ;)

Go read about synthetic division, it's very useful.

Synthetic division isn't all about guess and checking though (especially when you have a range of numbers using another theroem I can't remember, rational root theorem?).

In this case, you knew the roots, so synthetic division is perfect. Just remember that the divisor has to be in the form x-a.
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techno-freak
3 posts
#27 • 2 Y
Y by Adventure10, Mango247
the equation is x^2 +4x-5 roots r 1 and -5 concerning 2nd equation

2x^3+9x^2-6x-5=0 the sum of the roots is 1-5+p=4.5 p-4=4.5
therefore p=9.5 ( i may be mistaken though if p-4 =-4.5 then p=-0.5) :D :D
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pianoforte
1900 posts
#28 • 2 Y
Y by Kinzoku99, Adventure10
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ATGY
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#29
Y by
alternate solution
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IAmTheHazard
5000 posts
#30
Y by
*Casually revives 14-year-old thread*
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Overlord123
799 posts
#31
Y by
if you have a new solution to a problem, it is encouraged to share it.
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greenturtle3141
3537 posts
#32
Y by
...But it's not new...
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gabrupro
249 posts
#33 • 1 Y
Y by IAmTheHazard
The simplest solution:
Product of roots in cubic =$\frac{-5}{2}$
Product of roots in quadratic = $5$
Remaining root = $\frac{\frac{-5}{2}}{5}= \frac{-1}{2}$
This post has been edited 2 times. Last edited by gabrupro, Jun 20, 2020, 7:43 AM
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