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In the equation x^2+bx+12=0, x is a variable and b is a

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In the equation x^2+bx+12=0, x is a variable and b is a [#permalink] New post 21 Feb 2011, 04:38
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In the equation x^2+bx+12=0, x is a variable and b is a constant. What is the value of b?

(1) x-3 is a factor of x^2+bx+12.
(2) 4 is a factor of x^2+bx+12=0
[Reveal] Spoiler: OA

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Last edited by Baten80 on 21 Feb 2011, 05:46, edited 1 time in total.
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Re: QR:DS 62 Algebra Equation [#permalink] New post 21 Feb 2011, 05:18
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x^2+bx+12=0
a, coefficient of x^2=1
b, coefficient of x=b
c, constant in the quadratic polynomial=12

Let \alpha,\beta be the two factors of the equation.

(1) x-3 is a factor of x^2+bx+12.

\alpha=3
\alpha\beta=\frac{c}{a}=\frac{12}{1}
\beta=\frac{12}{3}=4

\alpha+\beta=\frac{-b}{a}=\frac{-b}{1}
b=-7
Sufficient.

(2) 4 is a factor of x^2+bx+12=0
Same as statement 1.

\beta=4
\alpha\beta=\frac{c}{a}=\frac{12}{1}
\alpha=\frac{12}{4}=3

\alpha+\beta=\frac{-b}{a}=\frac{-b}{1}
b=-7
Sufficient.

Ans: "D"
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Re: QR:DS 62 Algebra Equation [#permalink] New post 21 Feb 2011, 05:40
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Baten80 wrote:
In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b?
(1) x-3 is a factor of x^2+bx+12.
(2) 4 is a root of x^2+bx+12=0


You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.
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Re: QR:DS 62 Algebra Equation [#permalink] New post 21 Feb 2011, 05:56
Bunuel wrote:
Baten80 wrote:
In the equation x^2+bx+12=0, is variable and b is a constant. What is the value of b?
(1) x-3 is a factor of x^2+bx+12.
(2) 4 is a root of x^2+bx+12=0


You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.


Agree!!! This is much easier and way faster than the Viete's formula (I didn't know that's what it's called). Double thanks!!
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Re: QR:DS 62 Algebra Equation [#permalink] New post 21 Feb 2011, 05:59
Bunuel wrote:
Baten80 wrote:
In the equation x^2+bx+12=0, x is a variable and b is a constant. What is the value of b?
(1) x-3 is a factor of x^2+bx+12.
(2) 4 is a factor of x^2+bx+12=0


You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.


First post was wrong. This the correct post.
But I think answer is same.
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Re: QR:DS 62 Algebra Equation [#permalink] New post 21 Feb 2011, 06:16
We could also go by the quadratic extensiosn of the equation :
Since we know it has to have two rootsr Variables -

Statement 1 :-
X^2 - 3x-4x+12 = 0

x(x-3)-4(x-3) = 0

We have two root - X-3 , X-4 , So be has to be -3+-4 = -7 Sufficient :)

Statement 2:-

Since X = 4 it has be a root X-4, again using the same Expressions as above.

We get B=-7. Sufficient

Ans - D.
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In the equation ... [#permalink] New post 28 Feb 2011, 01:42
In the equation x^2+bx+12=0, x is a variable and b is a constant. What`s the value of b?

1) x-3 is a factor of x^2+bx+12
2) 4 is a root of the equation x^2 + bx + 12=0

Why (1) is sufficient?
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Re: In the equation ... [#permalink] New post 28 Feb 2011, 01:50
For equation x^2 + bx + 12 = 0, we need to find value of b

1) says x-3 is a factor of x^2+bx+12, or x=3 is one of the roots of this quadratic equation, so 3^2+3b+12 = 0 or 21+3b = 0 or b = -7. Sufficient
2) says 4 is one of the roots so 4^2+4b+12 = 0 or 28+4b = 0 or b = -7. Again sufficient, so D

if x-a is a factor of any polynomial in x, then at x=a, the value of such a polynomial would be 0 or in other words x = a is one of the roots of such a polynomial
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Re: In the equation ... [#permalink] New post 28 Feb 2011, 02:09
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Re: In the equation x^2+bx+12=0, x is a variable and b is a [#permalink] New post 15 Dec 2012, 22:40
knew that if x-3 is a factor then 3 could be substituted to get x

but did not know the meaning of root :lol: thanks..

Bunuel,
is it helpful to memorize viete's formula?
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Re: In the equation x^2+bx+12=0, x is a variable and b is a [#permalink] New post 03 Oct 2013, 04:49
D...
just find factors, already x-3, x-4 is another factor to get 12 and -3x-4x= -7,
this is possible becz x2+ bx +12 = 0, here b = +/-, to proof equation only is b= -7
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Re: QR:DS 62 Algebra Equation [#permalink] New post 24 Oct 2013, 04:53
Bunuel wrote:
Baten80 wrote:
In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b?
(1) x-3 is a factor of x^2+bx+12.
(2) 4 is a root of x^2+bx+12=0


You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.


Im not really following your first point here.

How do you now that x=3 is a root of the equation? If the equation can be factorized in (X+2)(X+6), isn't three then also a factor from the equation?
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Re: QR:DS 62 Algebra Equation [#permalink] New post 24 Oct 2013, 06:49
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waltiebikkiebal wrote:
Bunuel wrote:
Baten80 wrote:
In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b?
(1) x-3 is a factor of x^2+bx+12.
(2) 4 is a root of x^2+bx+12=0


You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.


Im not really following your first point here.

How do you now that x=3 is a root of the equation? If the equation can be factorized in (X+2)(X+6), isn't three then also a factor from the equation?


If (x+2)(x+6)=0, then x-3 is NOT a factor of the quadratics. The roots of (x+2)(x+6)=0 are x=-3 and x=-6.

We are told that x-3 is a factor of x^2+bx+12, thus (x-3)(x-k)=0, thus x=3 is one of the roots.

Hope it's clear.
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Re: In the equation x^2+bx+12=0, x is a variable and b is a [#permalink] New post 24 Oct 2013, 09:17
Thanks for your response, but I still don't get why x-3 is not a factor.
How can we now for sure that it is not a factor?
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Re: In the equation x^2+bx+12=0, x is a variable and b is a [#permalink] New post 24 Oct 2013, 09:36
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Re: In the equation x^2+bx+12=0, x is a variable and b is a   [#permalink] 24 Oct 2013, 09:36
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