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Consider the equation ; x 4 + ax 3 - 6x 2 + ax + 1 find parameter of a for which equation has two distict positive roots ?

Consider the equation ; x4  +  ax3  - 6x2  + ax + 1  
find parameter of a for which equation has two distict positive roots ?

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7 Answers

Akshay
185 Points
8 years ago
The answer is a is (-infinity,2).
Consider only positive side of x-axis for analysis, that is x>0.
At a=2, you get f(x)=x4+2x3-6x2+2x+1,
Differentiating with respect to x: f’(x)=4x3+6x2-12x+2.
When a=2, the minimum value of f(x)=0 at x=1. You can find this by putting f’(x)=0 for a=2.

When a>2, {f(x) when a>2} > {f(x) when a=2} , as the terms for a are positive. Hence, by increasing a you are increasing the value for f(x) for any x. Hence minimum value will be more than 0. Hence complex roots on positive side.

When a

Hence a

ATB.
Akshay
185 Points
8 years ago
Its not properly uploaded in previous comment.
 
The answer is a is (-infinity,2).
Consider only positive side of x-axis for analysis, that is x>0.
At a=2, you get f(x)=x4+2x3-6x2+2x+1,
Differentiating with respect to x: f’(x)=4x3+6x2-12x+2.
When a=2, the minimum value of f(x)=0 at x=1. You can find this by putting f’(x)=0 for a=2.

When a>2, {f(x) when a>2} > {f(x) when a=2} , as the coefficients for a are positive. Hence, by increasing a you are increasing the value for f(x) for any x. Hence minimum value will be more than 0. Hence complex roots on positive side.

When a0 and f(infinity)>0 , you will get two values at which f(x) cuts +ve x -axis. 

Hence a

ATB.
Akshay
185 Points
8 years ago
When a
Akshay
185 Points
8 years ago
When aHence a

0 , you will get two values at which f(x) cuts +ve x -axis. 
Akshay
185 Points
8 years ago
When a is less than 2, {f(x) when aHence a

0 , you will get two values at which f(x) cuts +ve x -axis. 
Akshay
185 Points
8 years ago
When a is less than 2, {f(x) when a(is less than)2} Hence a is less than 2.

0 , you will get two values at which f(x) cuts +ve x -axis. 
Akshay
185 Points
8 years ago
When a(is less than)2, {f(x) when a(is less than)2} (is less than) {f(x) when a=2}. At x=1, you will get f(x)(is less than)0. As f(0)(is less than)0 and f(infinity)>0 , you will get two values at which f(x) cuts +ve x -axis. 

Hence a less than 2.
Dont know why my comments are not uploading properly. 

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