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Question

The number of integral values of x satisfying -x2+10x-16<x-2 is


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Solution

Find the number of integral values of x:

The given expression is -x2+10x-16<x-2.

Simplify the following expression:
-x2+10x-16<x-2-x2+10x-160x2-10x+160x2-2x-8x+160(x-2)(x-8)0(x-2)0,or(x-8)0x2,8(1)

Take square on both sides of -x2+10x-16<x-2:
-x2+10x-16<x2-4x+42x2-14x+20>0x2-7x+10>0(x-5)(x-2)>0(x-5)>0,or(x-2)>0x>5,orx<2x-,2U5,(2)
The points 6,7 and 8 are those that lie within the interval of both (1) and (2).

Thus, x=6,7,8

Therefore, the number of integral values of x is 6,7,8.


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