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a2 + 7a < 0
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28 Mar 2019, 13:29
28 Mar 2019, 13:29
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\(a^2 + 7a < 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
a |
0 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: a2 + 7a < 0
[#permalink]
28 Mar 2019, 23:22
28 Mar 2019, 23:22
1
\(a^2 + 7a < 0\)
\(a(a+7) < 0\)
Here, a +ve number will always lead to left hand side of inequality being greater than the right hand side which cannot be the case, if you take a =0 then LHS = RHS however for a -ve number such as a =-1 LHS < RHS.
\(a(a+7) < 0\)
Here, a +ve number will always lead to left hand side of inequality being greater than the right hand side which cannot be the case, if you take a =0 then LHS = RHS however for a -ve number such as a =-1 LHS < RHS.
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Re: a2 + 7a < 0
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26 May 2021, 03:46
26 May 2021, 03:46
amorphous wrote:
\(a^2 + 7a < 0\)
\(a(a+7) < 0\)
Here, a +ve number will always lead to left hand side of inequality being greater than the right hand side which cannot be the case, if you take a =0 then LHS = RHS however for a -ve number such as a =-1 LHS < RHS.
\(a(a+7) < 0\)
Here, a +ve number will always lead to left hand side of inequality being greater than the right hand side which cannot be the case, if you take a =0 then LHS = RHS however for a -ve number such as a =-1 LHS < RHS.
I want to add little bit on it. if a(a+7) < 0 , then, the region of a would be -7<a<0, hence , we get the solution.
