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# Root

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Director
Joined: 12 Jun 2006
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31 Aug 2007, 11:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I picked #s here. Generally, I don't like doing this. Does anyone have a faster method?

If one root of the equation is 2x^2 + 3x - k= 0 is 6, what is the value of k?

90
42
18
10
-10
VP
Joined: 10 Jun 2007
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31 Aug 2007, 13:16
ggarr wrote:
I picked #s here. Generally, I don't like doing this. Does anyone have a faster method?

If one root of the equation is 2x^2 + 3x - k= 0 is 6, what is the value of k?

90
42
18
10
-10

90

2x^2 +3x - k = (x-6)*(2x-a) = 2x^2 - ax-12x + 6a
= 2x^2 + x(-a-12) + 6a

-a-12 = 3
a = -15

6a = -k
6*(-15) = -k
k = 90
VP
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31 Aug 2007, 13:20
2x^2 + 3x - k = 0

x = (-b ± sqrt(b^2-4*a*c))/2*a

x = (-3 ± sqrt(3^2-4*2*-k))/2*2

x = (-3 ± sqrt(9-8*-k))/4

one solution is given as 6 so:

6 = (-3 + sqrt(9-8*-k))/4

24 = -3 + sqrt(9-8*-k)

27*27 = 9-8*-k

720 = 8*k

k = 90

Senior Manager
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01 Sep 2007, 11:28
If 6 is the root, the equation should be satisfied for x= 6.

Putting x= 6 in equation.

2(6)^2 +3(6) - k = 0

72 + 18 - k =0

=> k = 90.

The ans is (A).

10 sec max
01 Sep 2007, 11:28
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