If x^2 + 4x + n 13 for all x, then which of the following must be tr

Question

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

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BrentGMATPrepNow · Answer

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

BrentGMATPrepNow · Answer

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

Kinshook · Answer

x^2 + 4x + n > 13 
  (x+2)^2 > 17-n 
If n=17; x=-2
But if n>17; 17-n <0; (x+2)^2> 17-n for all values of x

IMO A

Fdambro294 · Answer

Concept: the Graph of a Quadratic Expression will produce a Parabola in the coordinate plane in which the Input values (x) will produce corresponding Output values (y) such that the graph will be “U-shaped”

Step 1: change the quadratic expression into Vertex Form

(x)^2 + 4x + n > 13

(x)^2 + 4x + 4 - 4 + n > 13

(x + 2)^2 - 4 + n > 13

(x + 2)^2 - 17 + n > 0

step 2: analyze the Parabola

since the coefficient in front of the (x)^2 term is Positive, this will be an upwards opening parabola in which the Vertex will be the MINIMUM Point on the parabola.

Therefore, the minimum output value will be y = -17 at the coordinate point (-2 , -17)

In other words, the minimum output value from any corresponding X-input value will be ——-> -17

Therefore, to ensure that the output value is (+)positive, we need to Shift the Parabola upwards along the Y Axis

If we Add + 17 outside the square’s term, this will shift the parabola upwards such that the vertex will now fall on the X-axis. That means we can still get an output of 0

So to ensure that we always have an output of greater than > 0

We must shift the parabola up by a little more than 17 units

And to do that we need to insert a value into n that is greater than > 17

n > 17

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bumpbot · Answer

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If x^2 + 4x + n > 13 for all x, then which of the following must be tr

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