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x^3(x+1)=(x+k)(x+2k) where k belongs to (0.75,1) find- i)no. of real roots ii) greatest real root iii) least real root plz help me to solve this or guide me approaching this sum and more such types....... thanks in advance........

8 years ago

Answers : (2)

M Murali Krishna
33 Points
							
The no of roots of this equation are the no of points of intersection of y=x^4+x^3 and y=x^2+3kx+2k^2.
Differentiating y=x^4+x^3 wrt x we get dy/dx=4x^3+3x^2=x^2(4x+3). dy/dx>0 when x> -3/4, dy/dx
The graph of y=x^2+3kx+2k^2= (x-3k/2)^2-k^2/4 is a parabola with vertex at
(-3k/2,-k^2/4) By plotting these graphs we observe that there are two points of intersections. So the equation has two distinct real roots for each k belongs to (0.75,1) .
6 years ago
M Murali Krishna
33 Points
							
The no of roots of this equation are the no of points of intersection of y=x^4+x^3 and y=x^2+3kx+2k^2.
Differentiating y=x^4+x^3 wrt x we get dy/dx=4x^3+3x^2=x^2(4x+3). dy/dx>0 when x> -3/4, dy/dx
The graph of y=x^2+3kx+2k^2= (x-3k/2)^2-k^2/4 is a parabola with vertex at
(-3k/2,-k^2/4) By plotting these graphs we observe that there are two points of intersections. So the equation has two distinct real roots for each k belongs to (0.75,1) .
6 years ago
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