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#1 (permalink) |
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Trying to make mom and pop proud
Join Date: Jun 2002
Location: USA
Posts: 5
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If zē - 4z > 5 then which of the following is always true
A) z > -5 B) z < 5 C) z > -1 D) z < 1 E) z < -1 The equation can be written as zē - 4z - 5 > 0. . . (z - 5)(z + 1) > 0. If we set it equal to zero we find z > 5, z > -1. right?.... But the explanation given to this says z>5 and Z<-1 ..how is this?..I think i have some problem with solving these inequalities. Some one please help me. The aswer is E |
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#2 (permalink) |
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Trying to make mom and pop proud
![]() Join Date: Jul 2002
Posts: 25
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Hi Surabi,
The answer is right. It is indeed E. First of all, don't try to solve inequalities as if they are equalities as you have done. Now coming to the question, zē - 4z > 5 i.e. z(z-4)>5. Therefore either both z & z-4 are +ve or both are -ve(to be greater than 5) and the minimum value of z has to be at least 4 if z is +ve. z>4 is not an answer. Now if both are -ves, then the value of z must be less than -1 so that the product becomes more than 5. If you find yourself in a bind, always try to plug in some value & see the result. e.g. in this case if you take z>-1, suppose z=2. The result is 2(2-4) i.e.-4 which can't be greater than 5.I rest my case... HTH, ash ![]() |
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#4 (permalink) | |
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Within my grasp!
![]() ![]() Join Date: Sep 2002
Location: India
Posts: 164
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Quote:
There is a mathematical solution to such type of problems which is much easier to understand See zē - 4z > 5 => zē - 4z - 5 > 0 => (Z-5)(Z+1) > 0 This condition will hold true in two cases 1st case z-5 > 0 and Z + 1 > 0 => z > 5 and z > -1 means if z > 5 then z > -1 also so 1st case gives z > 5 2nd case both are less than 0 i.e. z-5 < 0 and z+1 < 0 i.e. z < 5 and z < -1 so if z < -1 it will be less than 5 also hence zē - 4z > 5 will be true if z > 5 or z < -1 And thats the answer DKP [^] |
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