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Question Stats:
54% (02:06) correct
46%
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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
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A) 16
B) 11
C) 9
D) 8
E) 7
Source: www.GMATinsight.com
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GMATinsight
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: u² + 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:
Cheers,
Brent
RUSDI
Joined: 02 Feb 2017
Last visit: 28 Dec 2018
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Concentration: Entrepreneurship, Strategy
Schools: Insead Sept '17
Schools: Insead Sept '17
Posts: 12
GMATinsight
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
General Discussion
It should be D. As the two values which suffice are 9 and -1 after solving the quadratic.
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sashiim20
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WE:Information Technology (Consulting)
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Posts: 608
GMATinsight
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...
GMATinsight
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.
Posts: 15
GMATinsight
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
Posts: 1,324
The first thing I spot is if you move 30 over and set the expression = 0
[x - 4]^2 + [x - 4] - 30 = 0
+6 and -5 are two roots that multiply to -30 and add to +1
(1st) let the absolute value expression = A
[X - 4] = 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
[X - 4] = -6
Or
[X - 4] = +5
Since the output of an absolute value expression can never be negative, the only valid root is when the Modulus expression = +5
[X - 4] = +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
[x - 4]^2 + [x - 4] - 30 = 0
+6 and -5 are two roots that multiply to -30 and add to +1
(1st) let the absolute value expression = A
[X - 4] = 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
[X - 4] = -6
Or
[X - 4] = +5
Since the output of an absolute value expression can never be negative, the only valid root is when the Modulus expression = +5
[X - 4] = +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
Posts: 7
GMATinsight
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
Joined: 12 Jan 2019
Last visit: 17 May 2026
Posts: 393
Location: India
Concentration: Strategy, Leadership
Schools: Oxford Said'25 (D)
GMAT 1: 660 Q47 V34
Products:
GMATinsight
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
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.
|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.
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
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
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