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Let S be the set of points \((a,b)\) with \(0 \le a,b \le 1\)  such that the equation \(x^4 + ax^3 - bx^2 + ax + 1 = 0\)
has at least one real root. Determine the area of the graph of S

Guest Dec 7, 2018
 #1
avatar+789 
+3

So  0<= a and b <= 1?

 

Or is it 0<= a <= 1 and 0<= b <= 1?

 #2
avatar+27237 
+2

As long as b >= 2 - 2a the equation has at least one real root.  the line b = 2 - 2a is a straight line going from b = 2 when a = 0, to b = 0 when a = 1.  Hence the area of b >= 2 - 2a is a triangle of base 1 and height 2; i.e. the area is (1/2)*1*2 = 1.

Alan  Dec 7, 2018
 #3
avatar+94183 
0

Hi Alan,

How did you work out that  b must be greater or equal to 2-2a  ?

Melody  Dec 7, 2018
edited by Melody  Dec 7, 2018
 #4
avatar+27237 
+1

I solved the quadratic in terms of a and b (well, ok, I used a piece of software to do so!). This resulted in an expression under a square root sign that meant if b>=2-2a, two of the four roots would be real.

Alan  Dec 7, 2018
 #5
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0

How can the height be 2?, doesnt\(0 \le a,b \le 1\)mean \(0 \le a \le 1\)and \(0 \le b \le 1\)? So it would have to be inside a 1 by 1 square?

Guest Dec 8, 2018
 #6
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0

Also, I got b<=-2a+2, can you check yours

EDIT: Nevermind, found my error

Guest Dec 8, 2018
edited by Guest  Dec 8, 2018
 #9
avatar+94183 
0

Thanks Alan.

Melody  Dec 8, 2018
 #7
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+1

So wouldn't it be 1/4?

Guest Dec 8, 2018
 #8
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+1

1/4 is the correct answer

Guest Dec 8, 2018
edited by Guest  Dec 8, 2018
 #10
avatar+27237 
+1

Correct.  I forgot about b having to be less than or equal to 1 at the end!

Alan  Dec 8, 2018

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