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Equations problem

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by ksh » Sat May 24, 2008 2:59 am
Plug x=-2
This gives y=4-2k-6, for a quadratic equation, y=0
hence, -2k=2
k=-1

the ans is b
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by chidcguy » Sat May 24, 2008 2:13 pm
ksh wrote:Plug x=-2
This gives y=4-2k-6, for a quadratic equation, y=0
hence, -2k=2
k=-1

the ans is b
I am with you until the y=4-2k-6 part

How come you picked the quadratic equation y=0? Why not y =1 or y =2 and we can get different K values.
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by molt_llest » Sat May 24, 2008 11:59 pm
ksh wrote:Plug x=-2
This gives y=4-2k-6, for a quadratic equation, y=0
hence, -2k=2
k=-1

the ans is b
I took y as a number, so I got 0=x^2+k*x-(6+y).
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by airan » Sun May 25, 2008 8:54 am
Another way of solving it ..
-2 is the root of y= x^2+k*x-6
Equation is
X^2 + kx- 6, --> Make its roots ..they will be -3 and 2,
put the value in the equation, both will satisfy the value -1.
Thanks
Airan
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by chidcguy » Sun May 25, 2008 10:24 am
molt_llest wrote:
ksh wrote:Plug x=-2
This gives y=4-2k-6, for a quadratic equation, y=0
hence, -2k=2
k=-1

the ans is b
I took y as a number, so I got 0=x^2+k*x-(6+y).
The reason why we have to substitute y=0 when we substitute x=-2 is because -2 is the root of the equation. A root of an equation is a value for which the equation ax ^2 + bx + C=0 7th grade definition which I forgot :(
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by molt_llest » Mon May 26, 2008 9:35 am
chidcguy wrote:
molt_llest wrote:
ksh wrote:Plug x=-2
This gives y=4-2k-6, for a quadratic equation, y=0
hence, -2k=2
k=-1

the ans is b
I took y as a number, so I got 0=x^2+k*x-(6+y).
The reason why we have to substitute y=0 when we substitute x=-2 is because -2 is the root of the equation. A root of an equation is a value for which the equation ax ^2 + bx + C=0 7th grade definition which I forgot :(
The question doen't say that Y=0, from my point of view next one is also a quadratic equation 0=x^2+k*x-(6+y), Why should I take y=0?
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by Stuart@KaplanGMAT » Mon May 26, 2008 9:55 am
I agree that the question doesn't demand that y=0 and is, therefore, ambiguous - where is the question from?

Please post all sources, so we know if we should take the question seriously.

Here's why I don't think y must be 0:

y= x^2+k*x-6

well, let's pick another value for y, say y = 8; we can just rewrite the equation as:

8 = x^2+k*x-6

0 = x^2+k*x-14

which is a solvable quadratic:

0 = (x-7)(x+2)

with roots of 7 and -2. With this value for y, k = -5.

So, the value of k depends on the value of y, making this questions unsolvable as written.
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