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Re: QR:DS 62 Algebra Equation [#permalink]
21 Feb 2011, 05:40
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This post received KUDOS
Expert's post
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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.
Re: QR:DS 62 Algebra Equation [#permalink]
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!! _________________
Re: QR:DS 62 Algebra Equation [#permalink]
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. _________________
Re: In the equation ... [#permalink]
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
Re: In the equation x^2+bx+12=0, x is a variable and b is a [#permalink]
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
Re: QR:DS 62 Algebra Equation [#permalink]
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?
Re: QR:DS 62 Algebra Equation [#permalink]
24 Oct 2013, 06:49
Expert's post
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.
Re: In the equation x^2+bx+12=0, x is a variable and b is a [#permalink]
27 Dec 2014, 06:50
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