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How many real solutions of the eqn 6x 2 - 77[x] + 147 = 0 , where [x] is the integral part of x.

How many real solutions of the eqn 6x2 - 77[x] + 147 = 0 , where [x] is the integral part of x.

Grade:12

1 Answers

AskIITian Expert Priyasheel - IITD
8 Points
11 years ago

6x^2 -77[x] +147 = 0 ...(i)

Consider the equation, 6x^2 -77x +147 = 0  ...(ii),
Roots of this equation are 7/3, 63/6.
[7/3]=2 and [63/6]=10
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Put [x]=2 in equation (i) and solve for x. x comes out to be sqrt(7/6) and [sqrt(7/6)]=1
so this is not a valid solution.
Again put [x]=10 in (i) and solve for x. x comes out to be sqrt(623/6) and [sqrt(623/6)]=10
so this is a valid solution.
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Other real solutions will lie between 2 and 10.
Substitute [x]=3, 4, 5, 6, 7,8, 9 one by one and solve for x and analyse each case as above.
You will find that for [x]=3, 9, 10, x comes out to be such, that [x (after solving)] = 3, 9, 10.
Hence the number of real solutions to the original equation will be 6. 

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