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# If a ≥ 0 and a =√8ab − 16b^2, then in terms of b, a =

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Joined: 02 Sep 2009
Posts: 62637
If a ≥ 0 and a =√8ab − 16b^2, then in terms of b, a =  [#permalink]

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04 Oct 2018, 00:45
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Difficulty:

35% (medium)

Question Stats:

76% (01:47) correct 24% (02:42) wrong based on 87 sessions

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If a ≥ 0 and $$a =\sqrt{8ab − 16b^2}$$, then in terms of b, a =

(A) −4b
(B) b/4
(C) b
(D) 4b
(E) 4b^2

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Re: If a ≥ 0 and a =√8ab − 16b^2, then in terms of b, a =  [#permalink]

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04 Oct 2018, 02:53
Bunuel wrote:
If a ≥ 0 and $$a =\sqrt{8ab − 16b^2}$$, then in terms of b, a =

(A) −4b
(B) b/4
(C) b
(D) 4b
(E) 4b^2

$$a =\sqrt{8ab − 16b^2}$$

Squaring both sides we get

$$a^2 =8ab − 16b^2$$
$$a^2 + 16b^2 - 8ab = 0$$

i.e. $$(a-4b)^2 = 0$$

i.e. $$a = 4b$$

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Re: If a ≥ 0 and a =√8ab − 16b^2, then in terms of b, a =  [#permalink]

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09 Feb 2020, 07:44
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Re: If a ≥ 0 and a =√8ab − 16b^2, then in terms of b, a =   [#permalink] 09 Feb 2020, 07:44
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