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If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ?

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Bunuel
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If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ? [#permalink] New post   30 Dec 2020, 21:30
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If \(x^2 = ax + 12a^2\), what are the 2 solutions for x in terms of a ?

(A) –6a and –2a
(B) –4a and –3a
(C) –4a and 3a
(D)–3a and 4a
(E) –2a and 6a
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D
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Re: If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ? [#permalink] New post   30 Dec 2020, 21:41
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Bunuel wrote:
If \(x^2 = ax + 12a^2\), what are the 2 solutions for x in terms of a ?

(A) –6a and –2a
(B) –4a and –3a
(C) –4a and 3a
(D)–3a and 4a
(E) –2a and 6a


\(x^2 = ax + 12a^2\)
\(x^2 - ax - 12a^2=0\)
\(x^2 - 4ax+3ax - 12a^2=0\)
\(x(x-4a)+3a(x-4a)=0\)
\((x+3a)(x-4a)=0\)
So, x can be 4a or -3a

D
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Re: If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ? [#permalink] New post   30 Dec 2020, 21:50
Asked: If \(x^2 = ax + 12a^2\), what are the 2 solutions for x in terms of a ?

xˆ2 - ax - 12aˆ2 = 0
xˆ2 - 4ax + 3ax - 12aˆ2 = 0
(x-4a)(x+3a)= 0
x = {4a, -3a}

IMO D
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Re: If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ? [#permalink] New post   30 Dec 2020, 23:32
\(x^2 = ax + 12a^2\)

\(x^2 - ax - 12a^2=0\)

\(x^2 - 4ax+3ax - 12a^2=0\)

\(x(x-4a)+3a(x-4a)=0\)

\((x+3a)(x-4a)=0\)

\(x= 4a, -3a\)

Ans D
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If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ? [#permalink] New post   01 Jan 2021, 07:01
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Bunuel wrote:
If \(x^2 = ax + 12a^2\), what are the 2 solutions for x in terms of a ?

(A) –6a and –2a
(B) –4a and –3a
(C) –4a and 3a
(D)–3a and 4a
(E) –2a and 6a


\(x^2 = ax + 12a^2\)
\(x^2 - ax - 12a^2=0\)
\(x^2 - 4ax+3ax - 12a^2=0\)
\(x(x-4a)+3a(x-4a)=0\)
\((x+3a)(x-4a)=0\)
Now,
\((x+3a)=0, x=-3a, or, (x-4a)=0, x=4a\)

The two solutions of x are \(-3a, \ or \ 4a\).

The answer is \(D\)
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Re: If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ? [#permalink] New post   18 Feb 2024, 08:11
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Re: If x^2 = ax + 12a^2, what are the 2 solutions for x in terms of a ? [#permalink]
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