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Which is larger, the sum of the roots of equation A or the

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Which is larger, the sum of the roots of equation A or the  [#permalink]

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New post 30 Jul 2011, 15:05
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A: x^2 + 6x - 40 = 0
B: x^2 + kx + j = 0

Which is larger, the sum of the roots of equation A or the sum of the roots of equation B?

(1) j = k

(2) k is negative
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Re: sum of roots  [#permalink]

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New post 30 Jul 2011, 18:15
1
dreambeliever wrote:
A: \(x^2\) + 6x - 40 = 0
B: \(x^2\) + kx + j = 0
Which is larger, the sum of the roots of equation A or the sum of the roots of equation B?

1. j = k
2. k is negative


Sum of roots of an equation:
\(ax^2+bx+c=0\)
Is
\(\frac{-b}{a}\)

So,

A. \(x^2 + 6x - 40 = 0\)
Sum of roots\(=\frac{-6}{1}=-6\)

B. \(x^2 + kx + j = 0\)
Sum of roots\(=\frac{-k}{1}=-k\)

We need to find
Larger of (-6, -k)

1. \(j=k\)
k can be 100 or -100.
Not Sufficient.

2. k<0
Thus, -k>0
Any positive number will always be greater than -6.
Thus, -k > -6
OR
Sum of roots of B > Sum of roots of A
Sufficient.

Ans: "B"
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Re: sum of roots  [#permalink]

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New post 31 Jul 2011, 22:17
fluke wrote:
dreambeliever wrote:
A: \(x^2\) + 6x - 40 = 0
B: \(x^2\) + kx + j = 0
Which is larger, the sum of the roots of equation A or the sum of the roots of equation B?

1. j = k
2. k is negative


Sum of roots of an equation:
\(ax^2+bx+c=0\)
Is
\(\frac{-b}{a}\)

So,

A. \(x^2 + 6x - 40 = 0\)
Sum of roots\(=\frac{-6}{1}=-6\)

B. \(x^2 + kx + j = 0\)
Sum of roots\(=\frac{-k}{1}=-k\)

We need to find
Larger of (-6, -k)

1. \(j=k\)
k can be 100 or -100.
Not Sufficient.

2. k<0
Thus, -k>0
Any positive number will always be greater than -6.
Thus, -k > -6
OR
Sum of roots of B > Sum of roots of A
Sufficient.

Ans: "B"



Funnily enough I could not move on this till clock ticked one minutes then I moved to shortcuts (lazy me :P ) and picked B (obviously I was not sure though!) but when I saw only this part- sum of root = -b/a (Fluke's explanation) I was like ok I have done it right :lol:
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Re: sum of roots  [#permalink]

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New post 01 Aug 2011, 07:03
Sum of roots = -b/a

A Sum of roots = -6
B sum of roots = -k

so the comparison is between -6 and -k , to see which one is greater

1. Not sufficient
j=k

k=2 => -k>-6 = > B>A
k=8 => -k<-6 =>B<A

2. Sufficient

As k is negative -k is always going to be positive
=> -k is always greater than -6
=> sum of roots of B is always greater than sum of roots of A.

Answer is B.
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Re: sum of roots  [#permalink]

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New post 13 Aug 2011, 08:28
fluke wrote:
dreambeliever wrote:
2. k<0
Thus, -k>0
Any positive number will always be greater than -6.
Thus, -k > -6
OR
Sum of roots of B > Sum of roots of A
Sufficient.

Ans: "B"


Fluke,

Can you please explain the part show in the question better? I didn't understand how you obtained a positive number
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Re: sum of roots  [#permalink]

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New post 13 Aug 2011, 12:29
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to petrifiedbutstanding i'll try:
if k is negative
thenthe equation looks like: X^2 -kx + j = 0

so the sum of roots is: (minus)- (b)/(a) or in our case -(minus) (-k)/1
which makes it: --(minus minus) k/1 or +k which is a positve number.

if k is a positive number it will allways be bigeer than -6
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Re: A: x^2 + 6x - 40 = 0 B: x^2 + kx + j = 0 Which is larger,  [#permalink]

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New post 14 May 2014, 16:34
I don't understand how j and k could be anything other than 4. Can anybody give me roots of this equation other than 2 that both add to and multiply to the same number?
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Re: A: x^2 + 6x - 40 = 0 B: x^2 + kx + j = 0 Which is larger,  [#permalink]

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New post 15 May 2014, 01:11
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Re: A: x^2 + 6x - 40 = 0 B: x^2 + kx + j = 0 Which is larger,   [#permalink] 15 May 2014, 01:11
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