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# Which of the following is a root of the equation 2x^2 − 20x = 48?

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Math Expert
Joined: 02 Sep 2009
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Which of the following is a root of the equation 2x^2 − 20x = 48?  [#permalink]

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24 Aug 2018, 00:42
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70% (01:10) correct 30% (01:27) wrong based on 84 sessions

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Which of the following is a root of the equation 2x^2 − 20x = 48?

A. -4
B. 2
C. 6
D. 8
E. 12

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Director
Joined: 18 Jul 2018
Posts: 902
Location: India
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Re: Which of the following is a root of the equation 2x^2 − 20x = 48?  [#permalink]

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24 Aug 2018, 00:45
2$$x^2$$ − 20x = 48 can be reduced to $$x^2$$-10x-24 = 0.

$$x^2$$-12x+2x-24 = 0

x(x-12) + 2(x-12) = 0

Either x = 12 or x = -2.

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Re: Which of the following is a root of the equation 2x^2 − 20x = 48?  [#permalink]

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24 Aug 2018, 02:54
Bunuel wrote:
Which of the following is a root of the equation 2x^2 − 20x = 48?

A. -4
B. 2
C. 6
D. 8
E. 12

$$2x^2 - 20 = 48$$

$$2(x^2 - 10x) = 48$$

$$x^2 - 10x - 48 = 0$$

$$x^2 - 12x + 2x -48 = 0$$
x( x - 12) +2 ( x - 12) = 0

(x - 12) ( x + 8) = 0

Either x -12= 0 or x =12,

or

x + 2 = 0 or x = -2.

Manager
Joined: 12 Sep 2017
Posts: 246
Re: Which of the following is a root of the equation 2x^2 − 20x = 48?  [#permalink]

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25 Feb 2019, 18:24
Afc0892 wrote:
2$$x^2$$ − 20x = 48 can be reduced to $$x^2$$-10x-24 = 0.

$$x^2$$-12x+2x-24 = 0

x(x-12) + 2(x-12) = 0

Either x = 12 or x = -2.

Hello Afc0892 !

Is it possible to solve it by Viete's theorem?

-b/a?
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Joined: 25 Feb 2019
Posts: 282
Re: Which of the following is a root of the equation 2x^2 − 20x = 48?  [#permalink]

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25 Feb 2019, 18:38
we can quickly put the value of root in equation and see which one satisfy this

12 is correct.

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Re: Which of the following is a root of the equation 2x^2 − 20x = 48?  [#permalink]

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02 Mar 2019, 10:10
Bunuel wrote:
Which of the following is a root of the equation 2x^2 − 20x = 48?

A. -4
B. 2
C. 6
D. 8
E. 12

Dividing the equation by 2, we have:

x^2 - 10x = 24

x^2 - 10x - 24 = 0

(x - 12)(x + 2) = 0

x = 12 or x = -2

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Re: Which of the following is a root of the equation 2x^2 − 20x = 48?   [#permalink] 02 Mar 2019, 10:10
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