Carcass wrote:
If x2+4x+n>13 for all x, then which of the following must be true ?
A. n > 17
B. n = 20
C. n = 17
D. n < 11
E. n = 13
Key concept: k² ≥ 0 for all values of kSo let's see if we can rewrite the left side of the inequality as the square of some value
So far, we have
x2+4xSince
x2+4x+4=(x+2)(x+2)=(x+2)2, let's...
Take:
x2+4x+n>13Add 4 to both sides to get:
x2+4x+n+4>17 Now subtract n from both sides to get:
x2+4x+4>17−nRewrite as follows:
(x+2)2>17−nIMPORTANT: It's tempting to conclude that n = 17 (answer choice C)
If n = 17, our inequality becomes
(x+2)2>0Notice that this inequality is NOT TRUE when
x=−2So, the correct answer is not C
However, if n > 17, then 17-n is
negative, which means the inequality will hold true for ALL values of x.
Answer: A
Cheers,
Brent