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 \(n = 2\).
Answer B is correct.
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