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Re: QR:DS 62 Algebra Equation [#permalink]
21 Feb 2011, 05:40

2

This post received KUDOS

Expert's post

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.

For my Cambridge essay I have to write down by short and long term career objectives as a part of the personal statement. Easy enough I said, done it...