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

8 Points
```										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____________________________________________________Put [x]=2 in equation (i) and solve for x. x comes out to be sqrt(7/6) and [sqrt(7/6)]=1so 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)]=10so this is a valid solution.____________________________________________________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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8 years ago
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