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If b and c are constants for which the quadratic equation x^2+bx+c=0

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If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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If b and c are constants for which the quadratic equation \(x^2+bx+c=0\) has two different roots, what is the product of the two roots?

(1) One of the roots is 3.

(2) c = 6
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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broilerc wrote:
If b and c are constants for which the quadratic equation \(x^2+bx+c=0\) has two different roots, what is the product of the two roots?

(1) One of the roots is 3.

(2) c=6


--

Can someone elaborate please?



Hi
for a quadratic equation \(ax^2+bx+c=0\)..
SUM of roots = \(-\frac{b}{a}\)and PRODUCT of roots =\(\frac{c}{a}\)..
here a is 1, so we just require 'c'..

statement II gives us value of 'c'..
hence suff
B
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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broilerc wrote:
If b and c are constants for which the quadratic equation \(x^2+bx+c=0\) has two different roots, what is the product of the two roots?

(1) One of the roots is 3.

(2) c=6


--

Can someone elaborate please?


Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus for x^2+bx+c=0 the product of the roots will be c/1 = c.

Answer: B.
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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If b and c are constants for which the quadratic equation \(x^2+bx+c=0\) has two different roots, what is the product of the two roots?

(1) One of the roots is 3.

(2) c=6


We are given that the quadratic equation x^2 + bx + c = 0 has two different roots. We need to determine the product of these two roots.

Recall that in a quadratic equation in the form ax^2 + bx + c = 0, the sum of the two roots is –b/a and the product of the two roots is c/a. Here, we are given the equation in the form of x^2 + bx + c = 0, and since a = 1, the product of the two roots is c/1 = c.

Thus, if we know the value of c, then we know the product of the two roots.

Statement One Alone:

One of the roots is 3.

We are given that one of the roots is 3. However, since we don’t know the value of the other root, we can’t determine the product of the two roots. Statement one alone is not sufficient. Eliminate choices A and D.

Statement Two Alone:

c = 6

Since we know the value of c, the product of the two roots must be 6 (see problem stem analysis above). Statement two alone is sufficient.

Answer: B
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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broilerc wrote:
If b and c are constants for which the quadratic equation \(x^2+bx+c=0\) has two different roots, what is the product of the two roots?

(1) One of the roots is 3.

(2) c=6


--

Can someone elaborate please?


When we factorize a quadratic equation, we factorize such that sum of two roots = b and product of two roots = c*a

Here because a=1, the product will depend on the c

B is the answer
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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New post 23 Oct 2016, 15:48
Could someone please tell me if my approach is correct?


If \(x^2 + bx +c = 0\)
Then, (x + p)(x + q) = 0

Therefore
pq = c
p + q = b

The roots of the quadratic equation will be -p and -q.
The question is asking about the product of the roots, meaning (-p)(-q) = ?
This is equal to c = ?, which in turn is equal to pq = ?

- Statement 1: Insufficient, since it only provides -p = 3 or -q = 3
- Statement 2: Sufficient.

OA = B
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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New post 07 Apr 2017, 22:07
Bunuel wrote:
broilerc wrote:
If b and c are constants for which the quadratic equation \(x^2+bx+c=0\) has two different roots, what is the product of the two roots?

(1) One of the roots is 3.

(2) c=6


--

Can someone elaborate please?


Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus for x^2+bx+c=0 the product of the roots will be c/1 = c.

Answer: B.


Hi Bunuel, is Viete's theorem needed for the GMAT? For this question can we not just deduce the answer?

1) If one root is 3, the other root could be anything. Not suff.

2) For positive c, both roots are either -ve or +ve and multiplied will give c.
In addition, for a negative c, one root will be +ve and one will be -ve. Again, multiplied together they will give c.

Suff. B
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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New post 31 Oct 2017, 12:38
For a quadratic equation, \(ax^2+bx+c=0\)
If we have two roots x1 and x2,
x1+x2 = \(-\frac{b}{a}\) and x1*x2 = \(\frac{c}{a}\)

We have been asked to find the value of x1*x2.
Given data: quadratic equation in hand that a=1

Coming to the statements :
1) One of the roots is 3.
Knowing one of the roots is not going to be enough, as the other root
can be any number and therefore, the product of the roots will be
different for every different value of the second root(Insufficient)
2) c=6
If a=1, the product of the roots is also c which is 6(Sufficient)(Option B)
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Re: If b and c are constants for which the quadratic equation x^2+bx+c=0 [#permalink]

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New post 26 Feb 2018, 00:43
broilerc wrote:
If b and c are constants for which the quadratic equation \(x^2+bx+c=0\) has two different roots, what is the product of the two roots?

(1) One of the roots is 3.

(2) c = 6

B is the answer as product of root is c/a and statement 2 provide us enough information whereas statement A is providing only one root which is insufficient.
Re: If b and c are constants for which the quadratic equation x^2+bx+c=0   [#permalink] 26 Feb 2018, 00:43
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