Quadratic Equations - Find Roots of Equation [on hold]

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If $x_1$, $x_2$ be the roots of the equation $x^2-3x+A=0$ and $x_3$, $x_4$ be those of equation $x^2-12x+B=0$, and $x_1$, $x_2$, $x_3$, $x_4$ be an increasing geometric progression, find $A$ and $B$.

linear-algebra abstract-algebra

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edited Aug 3 at 20:07

Javi

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asked Aug 3 at 18:57

Gaurav Choudhury

62

put on hold as off-topic by rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma Aug 3 at 19:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

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  A key part of the computation is to use Vieta's formulas for the case of a quadratic equation. Since the roots form a geometric progression $x_2=x_1r, x_3=x_1r^2$, and $x_4=x_1r^3$. Therefore, $x_1(1+r)=x_1+x_1r=x_1+x_2=3$ and $x_1(r^2+r^3)=x_1r^2+x_1r^3=x_3+x_4=12$. Therefore, $r^2=4$, or $r=pm 2$. It follows that $x_1=1$ or $x_1=-3$. You can compute $A=x_1x_2=x_1^2r$ and $B=x_3x_4=x_1^2r^5$ from that.
  – spiralstotheleft
  Aug 3 at 19:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since you only want an increasing progression, you can exclude the case $r=-2$, x_1=-3$, which will give a solution but that is not increasing.
  – spiralstotheleft
  Aug 3 at 19:14
  

  
  
  
  

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If $x_1$, $x_2$ be the roots of the equation $x^2-3x+A=0$ and $x_3$, $x_4$ be those of equation $x^2-12x+B=0$, and $x_1$, $x_2$, $x_3$, $x_4$ be an increasing geometric progression, find $A$ and $B$.

linear-algebra abstract-algebra

share|cite|improve this question

edited Aug 3 at 20:07

Javi

2,1381625

asked Aug 3 at 18:57

Gaurav Choudhury

62

put on hold as off-topic by rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma Aug 3 at 19:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A key part of the computation is to use Vieta's formulas for the case of a quadratic equation. Since the roots form a geometric progression $x_2=x_1r, x_3=x_1r^2$, and $x_4=x_1r^3$. Therefore, $x_1(1+r)=x_1+x_1r=x_1+x_2=3$ and $x_1(r^2+r^3)=x_1r^2+x_1r^3=x_3+x_4=12$. Therefore, $r^2=4$, or $r=pm 2$. It follows that $x_1=1$ or $x_1=-3$. You can compute $A=x_1x_2=x_1^2r$ and $B=x_3x_4=x_1^2r^5$ from that.
  – spiralstotheleft
  Aug 3 at 19:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since you only want an increasing progression, you can exclude the case $r=-2$, x_1=-3$, which will give a solution but that is not increasing.
  – spiralstotheleft
  Aug 3 at 19:14
  

  
  
  
  

add a comment | 

up vote
-1
down vote

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up vote
-1
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1

If $x_1$, $x_2$ be the roots of the equation $x^2-3x+A=0$ and $x_3$, $x_4$ be those of equation $x^2-12x+B=0$, and $x_1$, $x_2$, $x_3$, $x_4$ be an increasing geometric progression, find $A$ and $B$.

linear-algebra abstract-algebra

share|cite|improve this question

edited Aug 3 at 20:07

Javi

2,1381625

asked Aug 3 at 18:57

Gaurav Choudhury

62

If $x_1$, $x_2$ be the roots of the equation $x^2-3x+A=0$ and $x_3$, $x_4$ be those of equation $x^2-12x+B=0$, and $x_1$, $x_2$, $x_3$, $x_4$ be an increasing geometric progression, find $A$ and $B$.

linear-algebra abstract-algebra

share|cite|improve this question

edited Aug 3 at 20:07

Javi

2,1381625

asked Aug 3 at 18:57

Gaurav Choudhury

62

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 20:07

Javi

2,1381625

edited Aug 3 at 20:07

Javi

2,1381625

edited Aug 3 at 20:07

Javi

2,1381625

2,1381625

asked Aug 3 at 18:57

Gaurav Choudhury

62

asked Aug 3 at 18:57

Gaurav Choudhury

62

asked Aug 3 at 18:57

Gaurav Choudhury

62

62

put on hold as off-topic by rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma Aug 3 at 19:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma Aug 3 at 19:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – rschwieb, Mike Miller, amWhy, Xander Henderson, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A key part of the computation is to use Vieta's formulas for the case of a quadratic equation. Since the roots form a geometric progression $x_2=x_1r, x_3=x_1r^2$, and $x_4=x_1r^3$. Therefore, $x_1(1+r)=x_1+x_1r=x_1+x_2=3$ and $x_1(r^2+r^3)=x_1r^2+x_1r^3=x_3+x_4=12$. Therefore, $r^2=4$, or $r=pm 2$. It follows that $x_1=1$ or $x_1=-3$. You can compute $A=x_1x_2=x_1^2r$ and $B=x_3x_4=x_1^2r^5$ from that.
  – spiralstotheleft
  Aug 3 at 19:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since you only want an increasing progression, you can exclude the case $r=-2$, x_1=-3$, which will give a solution but that is not increasing.
  – spiralstotheleft
  Aug 3 at 19:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A key part of the computation is to use Vieta's formulas for the case of a quadratic equation. Since the roots form a geometric progression $x_2=x_1r, x_3=x_1r^2$, and $x_4=x_1r^3$. Therefore, $x_1(1+r)=x_1+x_1r=x_1+x_2=3$ and $x_1(r^2+r^3)=x_1r^2+x_1r^3=x_3+x_4=12$. Therefore, $r^2=4$, or $r=pm 2$. It follows that $x_1=1$ or $x_1=-3$. You can compute $A=x_1x_2=x_1^2r$ and $B=x_3x_4=x_1^2r^5$ from that.
  – spiralstotheleft
  Aug 3 at 19:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since you only want an increasing progression, you can exclude the case $r=-2$, x_1=-3$, which will give a solution but that is not increasing.
  – spiralstotheleft
  Aug 3 at 19:14
  

  
  
  
  

1

1

A key part of the computation is to use Vieta's formulas for the case of a quadratic equation. Since the roots form a geometric progression $x_2=x_1r, x_3=x_1r^2$, and $x_4=x_1r^3$. Therefore, $x_1(1+r)=x_1+x_1r=x_1+x_2=3$ and $x_1(r^2+r^3)=x_1r^2+x_1r^3=x_3+x_4=12$. Therefore, $r^2=4$, or $r=pm 2$. It follows that $x_1=1$ or $x_1=-3$. You can compute $A=x_1x_2=x_1^2r$ and $B=x_3x_4=x_1^2r^5$ from that.
– spiralstotheleft
Aug 3 at 19:08

A key part of the computation is to use Vieta's formulas for the case of a quadratic equation. Since the roots form a geometric progression $x_2=x_1r, x_3=x_1r^2$, and $x_4=x_1r^3$. Therefore, $x_1(1+r)=x_1+x_1r=x_1+x_2=3$ and $x_1(r^2+r^3)=x_1r^2+x_1r^3=x_3+x_4=12$. Therefore, $r^2=4$, or $r=pm 2$. It follows that $x_1=1$ or $x_1=-3$. You can compute $A=x_1x_2=x_1^2r$ and $B=x_3x_4=x_1^2r^5$ from that.
– spiralstotheleft
Aug 3 at 19:08

1

1

Since you only want an increasing progression, you can exclude the case $r=-2$, x_1=-3$, which will give a solution but that is not increasing.
– spiralstotheleft
Aug 3 at 19:14

Since you only want an increasing progression, you can exclude the case $r=-2$, x_1=-3$, which will give a solution but that is not increasing.
– spiralstotheleft
Aug 3 at 19:14

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2 Answers
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We have $$x_2=kcdot x_1$$ $$x_3=k^2cdot x_1$$ $$x_4=k^3cdot x_1$$ $$x_1+x_2=3$$ $$x_3+x_4=12$$

So, we have $(k+1)cdot x_1=3$ and $k^2(k+1)x_1=12$ giving $k=2$ and $x_1=1$. Hence we have $x_1=1,x_2=2,x_3=4,x_4=8$ giving $A=2$ , $B=32$

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answered Aug 3 at 19:10

Peter

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what i can see from seeing the above informations that u should try hard before seeing solution.anyway lets begin with the question.
we can assume x1=a, x2=ar, x3=ar^2, x4=ar^3
what we can use is the sum of roots .i.e. x1+x2=a(1+r)=3 and x3+x4=ar^2(1+r)=12
dividing both sums we have r^2=4 so r=2 or (-2)
for r=2 we have a=3/(1+2)=1 so x1=1,x2=2,x3=4,x4=8 thus A=x1*x2= 2 and B=x3*x4= 32
for r=-2 we have a=3/(1-2)=-3 so x1=-3,x2=6,x3=-12,x4=24 thus A=-18 and B=-288 computed as above
but as we have increasing gp so we ignore the case of r=-2
hope that it will help you!

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answered Aug 3 at 19:21

Anukul gaurav

795

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

We have $$x_2=kcdot x_1$$ $$x_3=k^2cdot x_1$$ $$x_4=k^3cdot x_1$$ $$x_1+x_2=3$$ $$x_3+x_4=12$$

So, we have $(k+1)cdot x_1=3$ and $k^2(k+1)x_1=12$ giving $k=2$ and $x_1=1$. Hence we have $x_1=1,x_2=2,x_3=4,x_4=8$ giving $A=2$ , $B=32$

share|cite|improve this answer

answered Aug 3 at 19:10

Peter

44.8k938119

add a comment | 

up vote
1
down vote

We have $$x_2=kcdot x_1$$ $$x_3=k^2cdot x_1$$ $$x_4=k^3cdot x_1$$ $$x_1+x_2=3$$ $$x_3+x_4=12$$

So, we have $(k+1)cdot x_1=3$ and $k^2(k+1)x_1=12$ giving $k=2$ and $x_1=1$. Hence we have $x_1=1,x_2=2,x_3=4,x_4=8$ giving $A=2$ , $B=32$

share|cite|improve this answer

answered Aug 3 at 19:10

Peter

44.8k938119

add a comment | 

up vote
1
down vote

up vote
1
down vote

We have $$x_2=kcdot x_1$$ $$x_3=k^2cdot x_1$$ $$x_4=k^3cdot x_1$$ $$x_1+x_2=3$$ $$x_3+x_4=12$$

So, we have $(k+1)cdot x_1=3$ and $k^2(k+1)x_1=12$ giving $k=2$ and $x_1=1$. Hence we have $x_1=1,x_2=2,x_3=4,x_4=8$ giving $A=2$ , $B=32$

share|cite|improve this answer

answered Aug 3 at 19:10

Peter

44.8k938119

We have $$x_2=kcdot x_1$$ $$x_3=k^2cdot x_1$$ $$x_4=k^3cdot x_1$$ $$x_1+x_2=3$$ $$x_3+x_4=12$$

So, we have $(k+1)cdot x_1=3$ and $k^2(k+1)x_1=12$ giving $k=2$ and $x_1=1$. Hence we have $x_1=1,x_2=2,x_3=4,x_4=8$ giving $A=2$ , $B=32$

share|cite|improve this answer

answered Aug 3 at 19:10

Peter

44.8k938119

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 19:10

Peter

44.8k938119

answered Aug 3 at 19:10

Peter

44.8k938119

answered Aug 3 at 19:10

Peter

44.8k938119

44.8k938119

add a comment | 

add a comment | 

up vote
0
down vote

what i can see from seeing the above informations that u should try hard before seeing solution.anyway lets begin with the question.
we can assume x1=a, x2=ar, x3=ar^2, x4=ar^3
what we can use is the sum of roots .i.e. x1+x2=a(1+r)=3 and x3+x4=ar^2(1+r)=12
dividing both sums we have r^2=4 so r=2 or (-2)
for r=2 we have a=3/(1+2)=1 so x1=1,x2=2,x3=4,x4=8 thus A=x1*x2= 2 and B=x3*x4= 32
for r=-2 we have a=3/(1-2)=-3 so x1=-3,x2=6,x3=-12,x4=24 thus A=-18 and B=-288 computed as above
but as we have increasing gp so we ignore the case of r=-2
hope that it will help you!

share|cite|improve this answer

answered Aug 3 at 19:21

Anukul gaurav

795

add a comment | 

up vote
0
down vote

what i can see from seeing the above informations that u should try hard before seeing solution.anyway lets begin with the question.
we can assume x1=a, x2=ar, x3=ar^2, x4=ar^3
what we can use is the sum of roots .i.e. x1+x2=a(1+r)=3 and x3+x4=ar^2(1+r)=12
dividing both sums we have r^2=4 so r=2 or (-2)
for r=2 we have a=3/(1+2)=1 so x1=1,x2=2,x3=4,x4=8 thus A=x1*x2= 2 and B=x3*x4= 32
for r=-2 we have a=3/(1-2)=-3 so x1=-3,x2=6,x3=-12,x4=24 thus A=-18 and B=-288 computed as above
but as we have increasing gp so we ignore the case of r=-2
hope that it will help you!

share|cite|improve this answer

answered Aug 3 at 19:21

Anukul gaurav

795

add a comment | 

up vote
0
down vote

up vote
0
down vote

what i can see from seeing the above informations that u should try hard before seeing solution.anyway lets begin with the question.
we can assume x1=a, x2=ar, x3=ar^2, x4=ar^3
what we can use is the sum of roots .i.e. x1+x2=a(1+r)=3 and x3+x4=ar^2(1+r)=12
dividing both sums we have r^2=4 so r=2 or (-2)
for r=2 we have a=3/(1+2)=1 so x1=1,x2=2,x3=4,x4=8 thus A=x1*x2= 2 and B=x3*x4= 32
for r=-2 we have a=3/(1-2)=-3 so x1=-3,x2=6,x3=-12,x4=24 thus A=-18 and B=-288 computed as above
but as we have increasing gp so we ignore the case of r=-2
hope that it will help you!

share|cite|improve this answer

answered Aug 3 at 19:21

Anukul gaurav

795

what i can see from seeing the above informations that u should try hard before seeing solution.anyway lets begin with the question.
we can assume x1=a, x2=ar, x3=ar^2, x4=ar^3
what we can use is the sum of roots .i.e. x1+x2=a(1+r)=3 and x3+x4=ar^2(1+r)=12
dividing both sums we have r^2=4 so r=2 or (-2)
for r=2 we have a=3/(1+2)=1 so x1=1,x2=2,x3=4,x4=8 thus A=x1*x2= 2 and B=x3*x4= 32
for r=-2 we have a=3/(1-2)=-3 so x1=-3,x2=6,x3=-12,x4=24 thus A=-18 and B=-288 computed as above
but as we have increasing gp so we ignore the case of r=-2
hope that it will help you!

share|cite|improve this answer

answered Aug 3 at 19:21

Anukul gaurav

795

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 19:21

Anukul gaurav

795

answered Aug 3 at 19:21

Anukul gaurav

795

answered Aug 3 at 19:21

Anukul gaurav

795

795

add a comment | 

add a comment |