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What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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21 Apr 2017, 06:46
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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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21 Apr 2017, 07:19
LHS = ( x+3 )^2 RHS is absolute value,,
square both RHS and LHS,, n den factorise.. im stuck then,,,any help plz



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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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What is the sum of all unique solutions for x2+6x+9=x+3x2+6x+9=x+3 ? A. 12 B. 9 C. 7 D. 5 E. 0 eq. 1 : x2 + 6x + 9 = x+3 x2 + 5x + 6 = 0 solves x = 3, 2. eq. 2 : x2 + 6x + 9 =  (x+3) x2 + 7x + 12 = 0 solves x = 4,3 Unique soln. 3, 4 ,2 Sum = 9. Ans. B
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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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21 Apr 2017, 07:34
mohshu wrote: LHS = ( x+3 )^2 RHS is absolute value,,
square both RHS and LHS,, n den factorise.. im stuck then,,,any help plz (x+3)^2 = x+3 if x>=3 (x+3)^2 = x+3 => x = 3 or 2 if x<3 (x+3)^2 = x3 => x=4 sum of all the roots of x =324=9



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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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22 Apr 2017, 02:09
What is the sum of all unique solutions for x^2+6x+9=x+3 ?
A. 12 B. 9 C. 7 D. 5 E. 0
x^2+6x+9= x+3 (Note  considering X+3 non negative only possible if X > 3) x^2+5x+6= 0 X= 2
x^2+6x+9= (x+3) x^2+7x+12= 0 X= 4 or 3
so My answer is B



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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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22 Apr 2017, 02:48
IMO B As LHS has inequality we have to take both positive and negative signs for LHS. Solving for that we will get sum of 9
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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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22 Apr 2017, 10:55
Bunuel wrote: What is the sum of all unique solutions for \(x^2+6x+9=x+3\) ?
A. 12 B. 9 C. 7 D. 5 E. 0 Here is my approach Eq. 1 x^2+6x+9 = x + 3 x^2+5x+6 = 0 (x+3)(x+2)=0 x1 = 3 x2=2 Eq. 2 x^2+6x+9 = x+3 x^2+7x+6 = 0 (x+1)(x+6) = 0 x1=1 x2=6 The sum = 3216=12 answer: (A) Please correct me if I'm wrong regards.



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What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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22 Apr 2017, 22:40
maxbm10 wrote: Bunuel wrote: What is the sum of all unique solutions for \(x^2+6x+9=x+3\) ?
A. 12 B. 9 C. 7 D. 5 E. 0 Here is my approach Eq. 1 x^2+6x+9 = x + 3 x^2+5x+6 = 0 (x+3)(x+2)=0 x1 = 3 x2=2 Eq. 2 x^2+6x+9 = x+3 x^2+7x+6 = 0 (x+1)(x+6) = 0 x1=1 x2=6 The sum = 3216=12 answer: (A) Please correct me if I'm wrong regards. I think you forgot the distribute the negative sign for X^2 +6x+9= (x+3) It should result in X^2+7x+12



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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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22 Apr 2017, 23:19
x^2+6x+9=x+3 (x+3)^2=x + 3 but (x+3)^2 ≥0 therefore, x^2+6x+9=(x+3) and x^2+6x+9 cannot be equal to (x+3) solving, x^2+6x+9=(x+3) will give us x=2 and x=3 therefore ans is x=5 option d



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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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27 Apr 2017, 18:16
Can someone please explain how the equation got to x^2+5x+6?



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What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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27 Apr 2017, 18:44
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x can take value either +x or x. We need to consider both the options if nothing is specified about x So considering two possibilties of the equation \(x^2\)+6x+9 = x+3 &&& \(x^2\)+6x+9=(x+3) \(x^2\)+6x+9  x  3 = 0 &&& \(x^2\)+6x+9 + x+3 = 0 \(x^2\)+5x+6=0 &&& \(x^2\)+7x+12=0 (x+3)(x+2)=0 &&& (x+3)(x+4) = 0 x=3,2 &&& x=3,4 Total Sum = 12 But we need unique solutions' sum so 3+2+4=9
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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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10 May 2017, 00:17
Bunuel wrote: What is the sum of all unique solutions for \(x^2+6x+9=x+3\) ?
A. 12 B. 9 C. 7 D. 5 E. 0 First, the lefthand side of the equation should be factored out: (x+3)(x+3)=lx +3l take the opposite value of the value inside each parentheses and add (3) + (3) (0)(0)= lx+3l only 3 can satisfy the equation so add 3 to 3 + 3 sum = 9 Hence "B"



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What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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10 May 2017, 02:55
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Bunuel wrote: What is the sum of all unique solutions for \(x^2+6x+9=x+3\) ?
A. 12 B. 9 C. 7 D. 5 E. 0 We have \(x^2+6x+9=(x+3)^2=x+3\) Solution 1.If \(x \geq 3 \implies x+3=x+3\) \( \implies (x+3)^2=x+3 \implies (x+3)^2(x+3)=0 \implies (x+3)(x+2)=0 \implies x =3 \, \, \, \text{or} \, \, \, x=2\) If \(x \leq 3 \implies x+3=(x+3)\) \( \implies (x+3)^2=(x+3) \implies (x+3)^2+(x+3)=0 \implies (x+3)(x+4)=0 \implies x =4 \) All roots for this equation are \(\{ 2, 3, 4 \}\). The sum of these roots are \(9\). The answer is B. Solution 2.\((x+3)^2=x+3 \iff x+3^2x+3=0 \iff x+3(x+31)=0 \implies x+3 =0 \, \, \text{or} \, \, x+3=1\) \(x+3=0 \implies x=3\) \(x+3=1 \implies x=2 \, \, \text{or} \, \, x=4\). The same result as Solution 1.
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Re: What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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12 May 2017, 12:37
Bunuel wrote: What is the sum of all unique solutions for \(x^2+6x+9=x+3\) ?
A. 12 B. 9 C. 7 D. 5 E. 0 We can solve the given absolute equation for when x + 3 is positive and for when x + 3 is negative. x + 3 is positive: x^2 + 6x + 9 = x + 3 x^2 + 5x + 6 = 0 (x + 2)(x + 3) = 0 x = 2 or x = 3 x + 3 is negative: x^2 + 6x + 9 = x  3 x^2 + 7x + 12 = 0 (x + 3)(x + 4) = 0 x = 3 or x = 4 Thus, the sum of the unique values of x is 2  3  4 = 9 Alternate Solution: We recognize that the right hand side is the expansion of (x + 3)^2; thus: (x + 3)^2 = x + 3 Let’s substitute x + 3 = u. Then, the equation becomes u^2 = u We are looking for numbers whose square is equal to its absolute value. This is only possible if u = 0, 1 or 1. Since u = x + 3, this means x + 3 = 0, x + 3 = 1 or x + 3 = 1. These equations give us the solutions x = 3, x = 2 and x = 4, respectively. Thus, the sum of the unique solutions is 3 + (2) + (4) = 9 Answer: B
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What is the sum of all unique solutions for x^2+6x+9=x+3 ? [#permalink]
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03 Aug 2017, 07:34
mrdlee23 wrote: Can someone please explain how the equation got to x^2+5x+6? When you have absolute values there are 2 equations since the absolute value can equal a negative or positive number. Essentially, the absolute value is a distance from the origin. So (X^2)+6X+9 = positive X+3 AND (X^2) +6X+9 = Negative X+3 So X^2+6X+9 = positive X+3 so that is (X^2) +6X+9=x+3 or (X^2)+5X+6=0 Solves into (x+3)(x+2) And AND X^2 +6X+9 = Negative X+3 = (X^2) +6X +9 = (X+3) or (X^2)+6X+9=x3 which equals (X^2) +7X+12. Solves into (X+3) and (x+4) So right now we are feeling pretty good we have solutions 4,3 and 2 however there could be one last trick so we have to ensure that the left side of the equation = a non negative value so check 4,3,2 in the equation (X^2)+6X+9 Notice the 3 solutions add to 1,0 & 1 so all equal a non negative value. The ABS value right side can be either negative or positive but the left side (in this case but I mean the non absolute value side) of the equation must be a non negative number. Why is that? Because the absolute value is a function which always shoots out a non negative value so 100 = 100 25 = 25 .100 = 100, 0 =0. So if the LHS of the equation equaled a negative number then it couldn't equal the RHS which is positive. That is the trick they didn't test here but it's important to note. One more comparison 100=100 but 100 does not = 100 as 100=100




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