What is the sum of all unique solutions for x^2+6x+9=|x+3| ?

Question

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

shekyonline · Accepted Answer

What is the sum of all unique solutions for x2+6x+9=|x+3|x2+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

mohshu · Answer

LHS = ( x+3 )^2 
RHS is absolute value,,

square both RHS and LHS,, n den factorise.. im stuck then,,,any help plz

attari92 · Answer

(x+3)^2 = |x+3| 
if x>=-3
(x+3)^2 = x+3 => x = -3 or -2
if x<-3
(x+3)^2 = -x-3 => x=-4
sum of all the roots of x =-3-2-4=-9

sreenu7464 · Answer

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

arvind910619 · Answer

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

maxbm10 · Answer

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 = -3-2-1-6=-12 answer: (A)

Please correct me if I'm wrong

regards.

Nunuboy1994 · Answer

I think you forgot the distribute the negative sign for 
X^2 +6x+9= -(x+3)

It should result in

X^2+7x+12

kapilsingal27 · Answer

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

mrdlee23 · Answer

Can someone please explain how the equation got to x^2 +5x+6 ?

umadurga · Answer

|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

Nunuboy1994 · Answer

First, the left-hand 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"

broall · Answer

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.

$(x+3)^2=|x+3| \iff |x+3|^2-|x+3|=0 \iff |x+3|(|x+3|-1)=0 \implies |x+3| =0 \, \, \text{or} \, \, |x+3|=1$
$|x+3|=0 \implies x=-3$
$|x+3|=1 \implies x=-2 \, \, \text{or} \, \, x=-4$.

The same result as Solution 1.

JeffTargetTestPrep · Answer

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

Samuelboyle96 · Answer

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=-x-3 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

plk · Answer

Hello experts!

I solved it in 2 ways: correct and incorrect. My mistake is:

Case1:
(x+3)^2=x+3
Case 2:
(x+3)^2=-1*(x+3)

it does attract to just divide both parts by (x+3) to get
x+3=1 and x+3=-1
But i am missing two solutions as i got rid of square. Please, help me to understand when it is okay to just devide both parts by something with x and not lose any solutuions. Thanks in advance!

anushkadhanwani · Answer

|x+3|>=0 so x>=-3

and when factors are taken out, we get -3,-4 and -2 shouldn't we ignore -4 since its <-3?

Vivek1707 · Answer

Notice that x^2+6+9 can be factorized to (x+3)^2

So the quesiton becomes (x+3)^2 = |x+3|

***Remember if a^2= |a| =, then a can have only the following 3 values i.e 1, -1 and 0 
If you take any other value for a and square it, It wont be equal to |a|

Similarly in this question there can be 3 cases i.e
x+ 3 = 1 i.e x = -2 
x+ 3 = -1 i.e x = -4
x+ 3 = 0 i.e x = -3

So if you add these up you get - 9

What is the sum of all unique solutions for x^2+6x+9=|x+3| ?

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What is the sum of all unique solutions for x^2+6x+9=|x+3| ?

Prep Toolkit

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