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Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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10 May 2015, 01:47

Bunuel 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 a correct solution.

Now, we can have 2 situations here Situation 1:x^2-8x+21=x-4+5 x^2-8x+21-x-5+4=x^2-9x+20 x^2-5x-4x+20 x=5 or x=4 Testing both values in the equation, both satisfy

Situation 2:x^2-8x+21=-x+4+5 x^2-7x+12=0 x^-3x-4x+12=0 x=3 or x=4 Testing both values in the equation, both satisfy Therefor sum of all solutions=3+4+5=12 Answer D

a. Unknowns: x Given: x2 – 8x + 21 = |x – 4| + 5. Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation. Question: What is the sum of all possible solutions for x?

b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer.

Attachment:

2015-05-11_1802.png [ 55.23 KiB | Viewed 9081 times ]

Sum of the different solutions: 5 + 4 + 3 = 12.

The answer is D.

Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right:

d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need not always be the case.
_________________

a. Unknowns: x Given: x2 – 8x + 21 = |x – 4| + 5. Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation. Question: What is the sum of all possible solutions for x?

b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer.

Attachment:

2015-05-11_1802.png

Sum of the different solutions: 5 + 4 + 3 = 12.

The answer is D.

Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right:

d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need not always be the case.

a. Unknowns: x Given: x2 – 8x + 21 = |x – 4| + 5. Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation. Question: What is the sum of all possible solutions for x?

b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer.

Attachment:

2015-05-11_1802.png

Sum of the different solutions: 5 + 4 + 3 = 12.

The answer is D.

Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right:

d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need not always be the case.

I have a conceptual doubt wrt such questions. Like most people above have, I also solved for the x>0 and x<0 cases and got the roots as 4,5 for x>0 and 3,4 for x<0. As we were considering case x<0 which doesn't get satisfied with the roots 3 and 4, I discarded those. Is that not right to do ?

Re: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| [#permalink]

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05 Jun 2017, 23:33

IMO D there will be two equations taking into account the two possible value of LHS i.e positive and negative . The solution we will have 3 different values of the solutions 3,4 and 5 Total =12

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.

My first thought here is that I want to separate out the absolute value. I see that I have a constant term on the left hand side too. So the first step is

a. Unknowns: x Given: x2 – 8x + 21 = |x – 4| + 5. Constraints: The question implies that there may be multiple solutions, as does the non-linear given equation. Question: What is the sum of all possible solutions for x?

b. Break into 2 equations (i.e. non-negative and negative inside absolute value sign) → Group all variables on one side of the equation, with 0 on the other → Factorable quadratic? → Factor → Solve → Answer.

Attachment:

2015-05-11_1802.png

Sum of the different solutions: 5 + 4 + 3 = 12.

The answer is D.

Alternatively, we could focus on the left side of the equation, which looks like a manageable quadratic, isolating the absolute value on the right:

d. On absolute value problems, some of the solutions might be duplicates. Just cross these off at the end. Additionally, we could check to make sure the results for each solutions satisfy the initial assumptions. For example, in Scenario 1 we assumed x – 4 ≥ 0. Both solutions, 5 and 4, satisfy this inequality. Similarly, the solutions 4 and 3 satisfy Scenario 2’s assumption, that x – 4 ≤ 0. Thus all the solutions were valid. For complex absolute value equations such as this one, this need not always be the case.

I have a conceptual doubt wrt such questions. Like most people above have, I also solved for the x>0 and x<0 cases and got the roots as 4,5 for x>0 and 3,4 for x<0. As we were considering case x<0 which doesn't get satisfied with the roots 3 and 4, I discarded those. Is that not right to do ?

Please help understand.

Thanks in advance! Shruti.

Responding to a pm:

The ranges for which you are solving are not x > 0 and x < 0. They are x >= 4 and x < 4 because

we say that |a| = a when a >= 0 |a| = -a when a < 0

So |x - 4| = (x - 4) when (x - 4) >= 0 i.e. x >= 4 |x - 4| = -(x - 4) when (x - 4) < 0 i.e. x < 4

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

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

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

(x-4)(x-4)=(x-4) (x-4) = (x-4)/(x-4) ------> (x-4)=1-----> x = 5

Scenario 2 where |x-4| is negative (x-4)(x-4)=-(x-4)---------> (x-4)=-(x-4)/(x-4)----------> (x-4)=-1--------> x = 3

So i have only two solutions x=5 and x=3...Why am I getting only 2 solutions ? while there are 3...

I know I am doing something very stupid with the basics. Can somebody please point out what is it ?

You cannot reduce (x-4)(x-4)=(x-4) by x-4 because x-4 can be 0 and we cannot divide by 0. By doing so you loose a root, namely x = 4.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. _________________