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If x^2 + 4x + n > 13 for all x, then which of the following
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14 Apr 2020, 08:31
14 Apr 2020, 08:31
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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
A. n > 17
B. n = 20
C. n = 17
D. n < 11
E. n = 13
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If x^2 + 4x + n > 13 for all x, then which of the following
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14 Apr 2020, 08:35
14 Apr 2020, 08:35
1
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
A. n > 17
B. n = 20
C. n = 17
D. n < 11
E. n = 13
Key concept: k² ≥ 0 for all values of k
So let's see if we can rewrite the left side of the inequality as the square of some value
So far, we have \(x^2 + 4x\)
Since \(x^2 + 4x +4 = (x+2)(x+2) = (x+2)^2\), let's...
Take: \(x^2 + 4x + n > 13\)
Add 4 to both sides to get: \(x^2 + 4x + n + 4 > 17\)
Now subtract n from both sides to get: \(x^2 + 4x + 4 > 17-n\)
Rewrite as follows: \((x+2)^2 > 17-n\)
IMPORTANT: It's tempting to conclude that n = 17 (answer choice C)
If n = 17, our inequality becomes \((x+2)^2 > 0\)
Notice that this inequality is NOT TRUE when \(x = -2\)
So, 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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If x^2 + 4x + n > 13 for all x, then which of the following
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14 Apr 2020, 08:36
14 Apr 2020, 08:36
1
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
A. n > 17
B. n = 20
C. n = 17
D. n < 11
E. n = 13
Alternate approach: number sense
Notice 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
