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# f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the

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Joined: 02 Sep 2009
Posts: 60645
f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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15 Dec 2019, 23:31
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55% (hard)

Question Stats:

52% (02:15) correct 48% (02:01) wrong based on 60 sessions

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Competition Mode Question

$$f(x) = x^2 + 4x + k = 12$$; $$f(-6) = 0$$; if k is a constant and n is the number for which $$f(n) = 0$$, what is the value of n?

A) 6
B) 2
C) 0
D) -2
E) -12

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Posts: 8336
Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 00:23
1
$$f(x) = x^2 + 4x + k = 12$$; $$f(-6) = 0$$; if k is a constant and n is the number for which $$f(n) = 0$$, what is the value of n?

We are already given as n=-6 as f(-6)=0, but let us look if there is any other value.

$$f(-6) = (-6)^2 + 4*(-6) + k = 0.........36-24+k=0....k=-12$$

$$f(n) = n^2 + 4n + k = 0......n^2+4n-12=0......n^2+6n-2n-12=0.....(n+6)(n-2)=0$$, so n=-6 or n=2

B
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Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 02:21
1
first determine value of k at f(-6) ; k = 0
check with f(x) = x^2+4x-12 =0
now the value of f(n)=0 we see comes
at n = 2
IMO B

f(x)=x2+4x+k=12f(x)=x2+4x+k=12; f(−6)=0f(−6)=0; if k is a constant and n is the number for which f(n)=0f(n)=0, what is the value of n?

A) 6
B) 2
C) 0
D) -2
E) -12
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Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 02:37
f(x)=x2+4x+k=12 , f(−6)=0 this results in k=0

so for f(x)=0, then x has to be '0'

OA:C
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Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 04:04
1
We are given that f(-6) = 0 which means that -6 is a solution of the function f(x).

f(-6) = (-6)^2 + 4*(-6) + k - 12 = 0.
from here, we get k = 0.

Now we need to find the value of n such that f(n) = 0, which means we need to find the second solution of f(x)

now, f(x) = x^2 + 4x - 12 = 0
so f(x) = (x+6)(x-2).

The other solution for f(x) = 2 = n such that f(n) or f(2) = 0

IMO, B
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Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 05:13
1
Quote:
Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the number for which f(n)=0, what is the value of n?

A) 6
B) 2
C) 0
D) -2
E) -12

f(x) = x^2 + 4x + k = 12
f(-6) = x^2 + 4x + k = 0, 36-24+k=0, k=-12
f(n) = x^2 + 4x + k = 0, x^2 + 4x + (-12) = 0,
(x+6)(x-2)=0, n=2

Ans (B)
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Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 06:38
1
f(x)=x^2+4x+k=12; f(-6)=0. We are to determine n such that f(n)=0
From the given information, f(-6)=0
f(x=-6)=36-24+k=0, k=-12
f(x)=x^2+4x-12.
To find n, we need to equate f(x)=0 and find the roots. the second root has to be n. We already know one root of f(x), x=-6.
x^2+4x-12=0
(x+6)(x-2)=0
So the roots of f(x) are -6, and 2.
n is therefore 2.

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Re: f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 13:43
1
$$f(x)=x^{2}+4x+k=12$$;
f(−6)=0;
--> if k is a constant and n is the number for which f(n)=0, what is the value of n?

$$f(−6)=x^{2}+4x+k -12=36 -24+ k- 12=0$$
--> k=0.

$$f(x)=x^{2}+4x-12=0$$;
--> $$f(n)= n^{2} +4n -12=0$$
(n+6)*(n-2)=0
--> n=-6 and n=2

Math Expert
Joined: 02 Sep 2009
Posts: 60645
f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the  [#permalink]

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16 Dec 2019, 23:51
Bunuel wrote:

Competition Mode Question

$$f(x) = x^2 + 4x + k = 12$$; $$f(-6) = 0$$; if k is a constant and n is the number for which $$f(n) = 0$$, what is the value of n?

A) 6
B) 2
C) 0
D) -2
E) -12

OFFICIAL EXPLANATION

There are two main approaches you can take to solving this problem. You can either solve for k and turn this problem into a simple factoring problem or you can ignore k and reverse factor.

Method 1: Solve for K.

$$f(x) = x^2 + 4x + k = 12$$
$$f(x) = x^2 + 4x + k - 12 = 0$$ (subtract 12)

Plug in $$x=-6$$ knowing that the value of f(x) must equal zero since $$f(-6)=0$$.
$$f(-6) = (-6)^2 + 4(-6) + k - 12 = 0$$
$$36 -24 + k - 12 = 0$$
$$0 - k = 0$$ --> $$k = 0$$

The equation is now:
$$f(x) = x^2 + 4x + 0 - 12 = 0$$
$$f(x) = x^2 + 4x - 12 = 0$$

By factoring: $$x^2 + 4x - 12 = 0$$ equals:
$$(x+6)(x+n)$$ Note: $$x+6$$ is from $$f(-6)=0$$

Since we need two numbers that add to +4 and multiply to -12, we know that +6 and -2 work. Consequently, (x-2) is a factor and therefore, $$f(+2)=0$$, so $$n = 2$$.

Method 2: Ignore K.
A crucial insight in unlocking this problem is recognizing that if f(a) = 0 and f(x) is a quadratic equation in the form $$x^2 + bx + c = 0$$, $$(x – a)$$ is a factor of the equation. In order to get the equation into quadratic form, subtract 12.
$$f(x) = x^2 + 4x + k = 12$$
$$f(x) = x^2 + 4x + k – 12 = 0$$

Since $$f(-6) = 0$$, $$(x+6)$$ is a factor.

It is important to remember how factoring works. Specifically, remember the following:
$$(x + d)(x + e) = x^2 + dex + de = 0$$
So: $$(x + 6)(x + a) = x^2 + 4x + (k – 12)$$

With this in mind, you know that $$6 + a = 4$$. So, $$a = -2$$ and therefore, $$(x - 2)$$ is the other factor of the quadratic. So, $$f(2) = 0$$ and $$n = 2$$.

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f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the   [#permalink] 16 Dec 2019, 23:51
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