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Given the equation x^2 + bx + c = 0

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Location: United Arab Emirates
Concentration: Finance, Entrepreneurship
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Given the equation x^2 + bx + c = 0  [#permalink]

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New post 02 Aug 2018, 23:54
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Given the equation x^2 + bx + c = 0, where b and c are constants, what is the value of c?

(1) The sum of the roots of the equation is zero.

(2) The sum of the squares of the roots of the equation is equal to 18.

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Re: Given the equation x^2 + bx + c = 0  [#permalink]

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New post 03 Aug 2018, 00:13
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(1) gives info about the roots that they have opposite +- signs. Not Suff.
(2) -\(3^2\)+\(3^2\)=18; \(3^2\)+\(3^2\)=18 c=-9 or c=9. Not Suff.

1-2 together. We know that the roots have opposite signs, so x=-3 x=3 c=-9. Suff.

IMO
Ans: C
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Given the equation x^2 + bx + c = 0  [#permalink]

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New post 03 Aug 2018, 08:23
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delta23 wrote:
Given the equation x^2 + bx + c = 0, where b and c are constants, what is the value of c?

(1) The sum of the roots of the equation is zero.

(2) The sum of the squares of the roots of the equation is equal to 18.


Given the equation \(x^2 + bx + c = 0\),
Sum of roots=\(\alpha+\beta\)=\(\frac{-b}{a}\)
Product of roots=\(\alpha*\beta\)=\(\frac{c}{a}\)
a=1

Question stem:- c=?

St1:-Sum of roots=\(\alpha+\beta\)=\(\frac{-b}{a}\)=0,
We are not in a position to say the value of c.
Insufficient.

St2:-\(\alpha^2+\beta^2\)=18
Or, \((\alpha+\beta)^2-2*\alpha*\beta=18\)
Or, \((\frac{-b}{a})^2-2*\frac{c}{a}=18\)
So, we can't determine the value of c since value of b is not known.
Insufficient.

Combining, we have
\(-2*\frac{c}{a}=18\)
Or, \(\frac{c}{1}=-9\)
Or, c=-9.

Ans. (C)
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Given the equation x^2 + bx + c = 0 &nbs [#permalink] 03 Aug 2018, 08:23
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