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# If 2a^2 + 0.5a - 1 = 0, then a(4a + 1) =

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Joined: 02 Sep 2009
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If 2a^2 + 0.5a - 1 = 0, then a(4a + 1) =  [#permalink]

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05 Sep 2017, 01:27
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15% (low)

Question Stats:

83% (01:42) correct 17% (02:31) wrong based on 57 sessions

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If $$2a^2 + 0.5a - 1 = 0$$, then $$a(4a + 1) =$$

A. 0.5
B. 2
C. 3.5
D. 4
E. 0

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If 2a^2 + 0.5a - 1 = 0, then a(4a + 1) =  [#permalink]

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05 Sep 2017, 01:30
We have been asked to find the value for $$a(4a + 1) = 4a^2 + a$$.

$$2a^2 + 0.5a - 1 = 0$$ when multiplied by 2, gives $$4a^2 + a - 2 = 0$$
Hence, the expression $$4a^2 + a - 2 = 0$$ can be re-written as $$4a^2 + a = 2$$(Option B)
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Re: If 2a^2 + 0.5a - 1 = 0, then a(4a + 1) =  [#permalink]

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05 Sep 2017, 01:31
Bunuel wrote:
If $$2a^2 + 0.5a - 1 = 0$$, then $$a(4a + 1) =$$

A. 0.5
B. 2
C. 3.5
D. 4
E. 0

- $$2a^2 +0.5a - 1 = 0$$, we can get rid of decimal by multiply by 2 -> $$4a^2 + a - 2 = 0$$.
- We can add both sides with 2 -> $$4a^2 + a = 2$$ -> $$4a^2 + a = -2$$ -> $$a(4a+1) = 2$$.

B.
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Re: If 2a^2 + 0.5a - 1 = 0, then a(4a + 1) =  [#permalink]

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07 Sep 2017, 16:47
Bunuel wrote:
If $$2a^2 + 0.5a - 1 = 0$$, then $$a(4a + 1) =$$

A. 0.5
B. 2
C. 3.5
D. 4
E. 0

2a^2 + 0.5a - 1 = 0

4a^2 + a - 2 = 0

4a^2 + a = 2

a(4a + 1) = 2

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Re: If 2a^2 + 0.5a - 1 = 0, then a(4a + 1) =   [#permalink] 07 Sep 2017, 16:47
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