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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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10 Mar 2021, 09:41
10 Mar 2021, 09:41
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Question Stats:
55% (01:46) correct
44% (02:26) wrong based on 98 sessions
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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
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A) 16
B) 11
C) 9
D) 8
E) 7
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: What is the sum of all the real values of x for which |x-4|^2 + |x-4|
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10 Mar 2021, 10:18
10 Mar 2021, 10:18
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Carcass wrote:
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
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A) 16
B) 11
C) 9
D) 8
E) 7
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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: D
Cheers,
Brent
General Discussion
Re: What is the sum of all the real values of x for which |x-4|^2 + |x-4|
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10 Mar 2021, 10:05
10 Mar 2021, 10:05
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take |x-4| = y
\(y^2 + y - 30\) = 0
(y+6)(y-5)=0
y = -6 / 5
|x-4| = -6 / 5
x-4 = 6 / -6 / 5 / -5
x = 10 / -2 / 9 / -1
It is given that the sum of real values of x,
X can't be 10 & -2 as by putting x in |x-4| = y we will not get -6 as an answer as mode is always +ve.
Possible values = 9 & -1 Sum = 8
Answer D
\(y^2 + y - 30\) = 0
(y+6)(y-5)=0
y = -6 / 5
|x-4| = -6 / 5
x-4 = 6 / -6 / 5 / -5
x = 10 / -2 / 9 / -1
It is given that the sum of real values of x,
X can't be 10 & -2 as by putting x in |x-4| = y we will not get -6 as an answer as mode is always +ve.
Possible values = 9 & -1 Sum = 8
Answer D
