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If the roots of the equation (a-b)x2 + (b-c) x+ (c - a)= 0 are equal. Prove that 2a=b+c.
Ans: (a-b)x2 + (b-c) x+ (c - a) = 0
T.P 2a = b + c
B2 - 4AC = 0
(b-c)2 - [4(a-b) (c - a)] = 0
b2-2bc + c2 - [4(ac-a2 - bc + ab)] = 0
⇒ b2-2bc + c2 - 4ac + 4a2 + 4bc - 4ab = 0
⇒ b2+ 2bc + c2 + 4a2 - 4ac - 4ab= 0
⇒ (b + c - 2a)2 = 0
⇒ b + c = 2a
Prove: cotA/2.cotB/2.cotC/2 = cotA/2+cotB/2+cotC/2
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the value of square root of 200multiplied by square root of 5=
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Find the middle term of the AP 1, 8, 15....505. A ns: Middle terms a + (n-1)d = 505 a + (n-1)7 = 505 n - 1 = 504/7 n = 73 ∴ 37th term is middle term a 37
Find the normalized differential equation which has {x, xex} as its fundamental set
What is algebra?
What is the greater of two consecutive negative integers whose product is 132? Let x = the lesser integer and let x + 1 = the greater integer. Because product is a key word for
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