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05 Sep 2017, 01:27
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B
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Difficulty:
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85% (01:27) correct
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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
A. 0.5
B. 2
C. 3.5
D. 4
E. 0
05 Sep 2017, 01:30
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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)
\(2a^2 + 0.5a - 1 = 0\) 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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05 Sep 2017, 01:31
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Bunuel
- \(2a^2 +0.5a - 1 = 0\), 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.
