Carcass wrote:
If \(x^2 + 4x + n > 13\) 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
Alternate approach: number senseNotice that, if n is a huge number (like \(1,000,013\)) then we get: \(x^2 + 4x + 1,000,013 > 13\)
Now subtract \(1,000,013\) from both sides of the inequality to get: \(x^2 + 4x > -1,000,000\)
Since \(x^2\) is always greater than or equal to zero, we can see that \(x^2 + 4x\) will almost always be positive.
The only time that \(x^2 + 4x\) is negative is when \(-4 < x < 0\), but even then \(x^2 + 4x\) is always
greater than \(-1,000,000\)
So, \(n = 1,000,013\) satisfies the given condition that \(x^2 + 4x + n > 13\) for all x
When we check the answer choices we see that:
- answer choice A allows for \(n\) to equal \(1,000,013\)
- answer choices B, C, D and E do NOT allow for \(n\) to equal \(1,000,013\)
Answer: A
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep