CBSE Class 10 Answered
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 | 24 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 | 25 Oct, 2012, 05:20: PM
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