Let f ( x ) be x⁴ - 4 x³ + a x² + b x + 1
If the zeroes of f ( x ) are positive real numbers find a and b
Plz answer the question
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RaquazaX:
Had the roots been integers , it would be direct by Rational root theoram XD
I learnt the theorem only today when u wrote the answer ..... so I used trial error
I too learnt the theoram after seeing your question only XD
XD
Answers
Answered by
8
let zeroes be x,y,z ,d
x+ y +z +d = 4
xyz + yzd + zdx + xyd = a
xy + yz + zd + dx + xz + yd = -b
xyzd = 1
take random numbers
as x + y + z + d= 4
xyzd = 1
as 1 = x= y = z= d
satisfy both
so take all zeroes as 1
xyz + yzd + zdx + xyd = a
4 = a
xy + yz + zd + dx + xz + yd = -b
6 = -b
b= -6
so a and b can be 4 and -6
x+ y +z +d = 4
xyz + yzd + zdx + xyd = a
xy + yz + zd + dx + xz + yd = -b
xyzd = 1
take random numbers
as x + y + z + d= 4
xyzd = 1
as 1 = x= y = z= d
satisfy both
so take all zeroes as 1
xyz + yzd + zdx + xyd = a
4 = a
xy + yz + zd + dx + xz + yd = -b
6 = -b
b= -6
so a and b can be 4 and -6
ooooooh by trial and error thank you :-)
Answered by
3
Let x1x1, x2x2, x3x3, x4x4 be the roots of the equation Sum of roots x1+x2+x3+x4=4x1+x2+x3+x4=4 Product of roots x1.x2.x3.x4=1x1.x2.x3.x4=1 We know AM > GM unless the quantities are equal Hence (x1+x2+x3+x4)/4=1>(x1.x2.x3.x4)1/4(x1+x2+x3+x4)/4=1>(x1.x2.x3.x4)1/4. Therefore x1=x2=x3=x4=1x1=x2=x3=x4=1
Thus x4−4x3+6x2−4x+1=0
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