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The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|
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11 Mar 2021, 03:51
11 Mar 2021, 03:51
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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 the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(A) –7
(B) 7
(C) 10
(D) 12
(E) 14
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: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|
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11 Mar 2021, 08:08
11 Mar 2021, 08:08
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Carcass wrote:
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 the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(A) –7
(B) 7
(C) 10
(D) 12
(E) 14
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots
Given: x² – 8x + 21 = |x – 4|+ 5
Subtract 5 from both sides: x² – 8x + 16 = |x – 4|
Apply above rule
We get two cases: x² – 8x + 16 = x – 4 and -(x² – 8x + 16) = x – 4
x² – 8x + 16 = x – 4
Add 4 to both sides: x² – 8x + 20 = x
Subtract x from both sides: x² – 9x + 20 = 0
Factor: (x - 5)(x - 4) = 0
So, x = 5 or x = 4
-(x² – 8x + 16) = x – 4
Simplify: -x² + 8x - 16 = x – 4
Add 4 to both sides: -x² + 8x - 12 = x
Subtract x from both sides: -x² + 7x - 12 = 0
Multiply both sides by -1 to get: x² - 7x + 12 = 0
Factor: (x - 3)(x - 4) = 0
So, x = 3 or x = 4
We have three potential solutions: x = 5, x = 4 and x = 3
Now let's test for EXTRANEOUS ROOTS
x = 5
Plug into original equation to get:
5² – 8(5) + 21 = |5 – 4|+ 5
Evaluate: 6 = 6
WORKS! So, x = 5 IS a valid solution
x = 4
Plug into original equation to get:
4² – 8(4) + 21 = |4 – 4|+ 5
Evaluate: 5 = 5
WORKS! So, x = 4 is a valid solution
x = 3
Plug into original equation to get:
3² – 8(3) + 21 = |3 – 4|+ 5
Evaluate: 6 = 6
WORKS! So, x = 3 is a valid solution
SUM of all solutions = 5 + 4 + 3 = 12
Answer: D
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The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4|
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23 Dec 2021, 09:40
23 Dec 2021, 09:40
1
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
