What is the sum of all the real values of x for which |x-4|^2 + |x-4|

Question

What is the sum of all the real values of x for which |x-4|^2 + |x-4| = 30? A) 16 B) 11 C) 9 D) 8 E) 7 Source: www.GMATinsight.com

BrentGMATPrepNow · Accepted Answer

To make things easier, we'll use a technique known as  u-substitution

Let  u = |x - 4|

So,  |x-4| ² +  |x-4|  = 30...
...becomes:  u ² +  u  = 30
Set this quadratic equal to zero: u² + u - 30 = 0
Factor: (u + 6)(u - 5) = 0
So, either u = -6 or u = 5

Since  u = |x - 4| , we can now that that either |x - 4| = -6 or |x - 4| = 5

Let's examine each case.

|x - 4| = -6
Since the absolute value is always greater than or equal to 0, it's IMPOSSIBLE for |something| = -6 
So, this equation has no solution

|x - 4| = 5
This means that x - 4 = 5 or x - 4 = -5
If x - 4 = 5, then  x = 9 
If x - 4 = -5, then  x = -1

So, the SUM OF THE SOLUTIONS =  9  + ( -1 ) = 8

Answer:  D

Cheers,
Brent

RUSDI · Answer

Assume y=|x-4| y^2+y-30=0 (y+6)(y-5)=0 y=-6 or y =5 For y=-6. |x-4|=-6 If x-4>=0 , x-4=-6, thus x=-2 (not ok) If x-4<0 , x-4 = 6. thus x=10 (not ok) For y=5. |x-4|=5 If x-4>=0 , x-4=5, thus x=9 (ok) If x-4<0 , x-4 =-5. thus x=-1 (ok) Sum of x = 9+(-1) = 8 (D) Kudos if it helps

BeingHan · Answer

It should be D. As the two values which suffice are 9 and -1 after solving the quadratic.

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sashiim20 · Answer

Checking options, i got Answer C. x = 9.

|x-4|^2  + |x-4| = 30
 |9-4|^2  + |9 - 4| =  5^2  + 5 = 25 + 5 = 30
Therefore x = 9. Answer C...

LeoN88 · Answer

Let |x-4| = t, so t>=0

Now t^2 + t -30=0

solving t= 5/-6
T can only assume non negative value, so...

|x-4|=5.
x can be -1 and 9.

Sum of all real values= 9-1=8.

rishabh195 · Answer

But why can't the answer be C. Upon entering the value 9 in x, we get 25+5=30.

Then why not C? And what's wrong with my method? Please throw some light.

Posted from my mobile device

Fdambro294 · Answer

The first thing I spot is if you move 30 over and set the expression = 0

^2 +   - 30 = 0

+6 and -5 are two roots that multiply to -30 and add to +1

(1st) let the absolute value expression = A

= A everywhere the expression appears

(A)^2 + A - 30 = 0

(A + 6) (A - 5) = 0

A = -6 or A = +5

(2nd) now solve for the Modulus

Either

= -6

Or

= +5

Since the output of an absolute value expression can never be negative, the only valid root is when the Modulus expression = +5

= +5

Translation: “on the real number line, X is exactly 5 units away from +4”

X = -1 or +9

Sum: 9 - 1 =

8
Answer (D)

Posted from my mobile device

TanviWatts · Answer

For questions on absolute values, you should always check the value obtained in the equation. 
For example, after solving the above, we get 2 equations,

x^2 - 7x -18 = 0 and x^2 -9x -10 = 0 
This gives x as -2, 9, -1 and 10 
But if we plug these values back into the actual equation, 9 and -1 work while the others don't equate. 
The real values are 9 and -1 and the sum is 8 D

Please give kudos if you like, stuck at 99

Thanks

udaypratapsingh99 · Answer

Ley |x-4| = y (where y is always positive) ......(a)

Equation becomes, 
 y^2 + y = 30 
=>  y^2 + y - 30 = 0 
=>  y=5  (Ignoring the negative value as y cannot be negative) .....(b)

From (a) and (b),
 |x-4| = 5
=> x-4 = +/- 5
=>x = -1,9

thus, the sum of all the real values of x for which  |x-4|^2 + |x-4| = 30 is -1+9 = 8

D is Correct

martinfzb · Answer

another way to solve this question:
|x-4|>0 must be true, let it equals to y
then x-4= +y and x-4=-y
therefore 
x=+y+4 and x=-y+4
so add them up to get the sum of x values 
+y+4-y+4= 8
so choose D.

ZIX · Answer

Here's a simpler approach:

Let |x-4| = a

Then,  a^2 + a = 30 
 a(a+1) = 30

What two consecutive numbers when multiplied together will result in 30? 5*6 and -5*-6, correct!

Therefore, a = 5 or -5, and (x-4) = 5 or -5. Solve and we get x = 9 or -1. 9 + (-1) = 8

bumpbot · Answer

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What is the sum of all the real values of x for which |x-4|^2 + |x-4|

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