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09 Jul 2020, 00:06
Kudos
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Bunuel
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:
(A) –7
(B) 7
(C) 10
(D) 12
(E) 14
Kudos for a correct solution.
(A) –7
(B) 7
(C) 10
(D) 12
(E) 14
Kudos for a correct solution.
Asked: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|+ 5 is equal to:
x^2 – 8x + 21 = |x – 4|+ 5
x^2 - 8x - |x-4| + 16 = 0
Case 1: x <= 4
x^2 - 8x + x - 4 + 16 = 0
x^2 -7x + 12 = 0
(x - 4)(x-3) = 0
x = 3 or 4
Case 2: x>4
x^2 - 8x + 4 - x + 16 = 0
x^2 -9x + 20 = 0
(x - 4)(x-5) = 0
x = 5 or 4
Combining 2 cases:
x = 3 or 4 or 5
Sum of all solutions of x = 3 + 4 + 5 = 12
IMO D
23 Dec 2021, 10:40
Kudos
Bookmarks
Given that \(x^2\) – 8x + 21 = |x – 4|+ 5 and we need to find the value of x
Let's open |x-4|. We can do that by taking two cases
Case 1: x-4 >= 0 or x >= 4
=> |x-4| = x-4 (Check this video to know how to deal with Absolute Value problems)
=>\( x^2\) – 8x + 21 = x – 4 + 5 = x + 1
=>\(x^2 \) - 8x - x + 21 - 1 = 0
=> \(x^2 \) - 9x + 20 = 0
=> \(x^2 \) - 5x - 4x + 20 = 0
=> x (x-5) - 4(x-5) = 0
=> (x-5)*(x-4) = 0
=> x = 4 or 5
Since, x>=4 is the condition so both 4 and 5 are possible values of x
Case 2: x-4 < 0 or x < 4
=> |x-4| = -(x-4) = -x + 4 (Check this video to know how to deal with Absolute Value problems)
=>\( x^2\) – 8x + 21 = -x + 4 + 5 = -x + 9
=>\(x^2 \) - 8x + x + 21 - 9 = 0
=> \(x^2 \) - 7x + 12 = 0
=> \(x^2 \) - 3x - 4x + 12 = 0
=> x (x-3) - 4(x-3) = 0
=> (x-3)*(x-4) = 0
=> x = 3 or 4
Since, x < 4 is the condition so x = 3 is the solution from this one
=> Possible values of x are 3 ,4 , 5
=> Sum of solutions = 3 + 4 + 5 = 12
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
Let's open |x-4|. We can do that by taking two cases
Case 1: x-4 >= 0 or x >= 4
=> |x-4| = x-4 (Check this video to know how to deal with Absolute Value problems)
=>\( x^2\) – 8x + 21 = x – 4 + 5 = x + 1
=>\(x^2 \) - 8x - x + 21 - 1 = 0
=> \(x^2 \) - 9x + 20 = 0
=> \(x^2 \) - 5x - 4x + 20 = 0
=> x (x-5) - 4(x-5) = 0
=> (x-5)*(x-4) = 0
=> x = 4 or 5
Since, x>=4 is the condition so both 4 and 5 are possible values of x
Case 2: x-4 < 0 or x < 4
=> |x-4| = -(x-4) = -x + 4 (Check this video to know how to deal with Absolute Value problems)
=>\( x^2\) – 8x + 21 = -x + 4 + 5 = -x + 9
=>\(x^2 \) - 8x + x + 21 - 9 = 0
=> \(x^2 \) - 7x + 12 = 0
=> \(x^2 \) - 3x - 4x + 12 = 0
=> x (x-3) - 4(x-3) = 0
=> (x-3)*(x-4) = 0
=> x = 3 or 4
Since, x < 4 is the condition so x = 3 is the solution from this one
=> Possible values of x are 3 ,4 , 5
=> Sum of solutions = 3 + 4 + 5 = 12
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
21 Oct 2024, 00:24
Kudos
Bookmarks
Aren’t we supposed to check for extraneous roots as well as part of the process?
Because y^2 is positive does not mean y needs to be positive. Because |y| is positive, doesn't mean y needs to be positive too.
If y is -5, y^2 = 25 (positive) and |y| = 5 (positive) but y is still negative.
In above, (x - 4)^2 and |x - 4| are positive but x - 4 may not be positive. It could be positive or negative.
KarishmaB
rsrighosh
Bunuel egmat VeritasKarishma GMATNinja
I have a conceptual question here
If it says
\((x-4)^2 = |x-4|\)
Doesnt this mean that \(x-4 >= 0\) all the time? (as \((x-4)^2\) is always positive)
Why are we considering\( x-4\) can be negative?
Can anybody please explain if my understanding is wrong.
Quote:
x^2 – 8x + 21 = |x – 4| + 5
x^2 – 8x + 16 = |x – 4|
(x – 4)(x – 4) = |x – 4|
(x – 4)2 = |x – 4|
“Something squared equals its absolute value. 1 squared equals 1, 0 squared equals 0...hmm, –1 squared equals 1...”
x – 4 = –1, 0, or 1
x = 3, 4, or 5
x^2 – 8x + 16 = |x – 4|
(x – 4)(x – 4) = |x – 4|
(x – 4)2 = |x – 4|
“Something squared equals its absolute value. 1 squared equals 1, 0 squared equals 0...hmm, –1 squared equals 1...”
x – 4 = –1, 0, or 1
x = 3, 4, or 5
I have a conceptual question here
If it says
\((x-4)^2 = |x-4|\)
Doesnt this mean that \(x-4 >= 0\) all the time? (as \((x-4)^2\) is always positive)
Why are we considering\( x-4\) can be negative?
Can anybody please explain if my understanding is wrong.
Because y^2 is positive does not mean y needs to be positive. Because |y| is positive, doesn't mean y needs to be positive too.
If y is -5, y^2 = 25 (positive) and |y| = 5 (positive) but y is still negative.
In above, (x - 4)^2 and |x - 4| are positive but x - 4 may not be positive. It could be positive or negative.
