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When 3x^2 + nx + 5 = 0 has only one solution, n=?

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When 3x^2 + nx + 5 = 0 has only one solution, n=?  [#permalink]

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New post 17 Mar 2019, 10:54
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  25% (medium)

Question Stats:

72% (01:18) correct 28% (02:04) wrong based on 64 sessions

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When \(3x^2 + nx + 5 = 0\) has only one solution, n=?

A) \(\sqrt{48}\)
B) \(\sqrt{80}\)
C) \(10\)
D) \(3\sqrt{2}\)
E) \(\sqrt{60}\)
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Re: When 3x^2 + nx + 5 = 0 has only one solution, n=?  [#permalink]

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New post 17 Mar 2019, 11:00
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If it has only one root, \(b^2 - 4ac = 0\). Hence, \(n^2 - 4*3*5 = 0\); \(n = \sqrt{60}\).
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Re: When 3x^2 + nx + 5 = 0 has only one solution, n=?  [#permalink]

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New post 21 Jul 2019, 22:33
lucajava wrote:
If it has only one root, \(b^2 - 4ac = 0\). Hence, \(n^2 - 4*3*5 = 0\); \(n = \sqrt{60}\).


Is this just basic application of the quadratic formula? It looks different than the quadratic I normally see.
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Re: When 3x^2 + nx + 5 = 0 has only one solution, n=?   [#permalink] 21 Jul 2019, 22:33
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When 3x^2 + nx + 5 = 0 has only one solution, n=?

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