q1 if the roots of the equation (c^-ab)x^-2(a^-bc)x+b^-ac=0 are equal prove that either a=0 or a cube+b cube+c cube=3abc? q2 a polygon of n sides has n(n-3)/2 diagonals how many sides has a polygon with 54 diagonals? q3 if sin alpha and cos alpha are thr roots of the equation ax^+bx+c=0 then provethat a^+2ac=b^? q4 if the roots of the equation (a^+b^)x^-2(ac+bd)x+(c^+d^)=0 are equal then prove that a/b=c/d?

Asked by arnav sharma | 24th Oct, 2012, 06:13: PM

Expert Answer:

Please ask one question at a time. The answer to your first two questions is given below.
 

1) We know that for real and equal roots D=0

so, b2-4ac = 0

{-2(a2-bc)}2 - 4 (c2-ab) (b2-ac) = 0

4(a2-bc)2 - 4(c2-ab)(b2-ac)=0

4( a4+b2c2-2a2bc) = 4(b2c2-ac3-ab3+a2bc)

a4+ac3+ab3 = 2a2bc+a2bc

a (a3+c3+b3) = 3a2bc

Hence a = 0 and a3+b3+c3 = 3abc

2) a polygon of n sides has n(n-3)/2 diagonals

For the required polygon, number of diagonals = 54

So, n(n-3)/2 = 54

n2 - 3n - 108 = 0

n2 - 12n + 9n - 108 = 0

(n - 12) (n + 9) = 0

Since, n cannot be negative, n = 12

So, the number of sides of the required polygon is 12.

Answered by  | 25th Oct, 2012, 05:20: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.