What is the product of all the solutions of x^2 + 4x + 7 = |x + 2| + 3

x^2+4x+7=|x+2|+3

x^2+4x+4=|x+2|

(X+2)^2 = |x+2|
Only 3 values possible,
x = -3, x = -2, x =-1
So multiplication = -6

Similar question to practice: what-is-the-product-of-all-the-solutions-of-x-2-4x-147152.html

Hi viktorija,

The "tricky" part of this question is the absolute value. When that type of math is involved, Test Takers often miss out on some of the possible answers, so it's important to do whatever you have to do to be sure that you've found all of the possibilities. To that end, I'm going to use ALL of the little clues that question gives me and avoid as much "crazy" math as possible.

We're given X^2+4X+7=|X+2|+3 and we asked for the product of ALL of the solutions.

Notice how the question asks us to find ALL of the solutions and NOT "both" solutions. This gets me thinking that there's probably more than 2 solutions. Also, quadratic equations usually have 1 or 2 solutions and absolute values usually have 1 or 2 solutions, so I'd be looking for up to 3 or 4 possibilities....

First, we simplify:

X^2+4X+7 = |X+2|+3
X^2+4X+4 = |X+2|
(X+2)(X+2) = |X+2|

Now, looking at that last line, there's NO WAY that x can be positive. Try TESTing any positive number - the two sides will NEVER equal. So there are NO positive answers.

Looking at the answer choices, they're all within a relatively small distance from 0, so the individual values for X are probably all relatively close to 0 as well. Let's find them with a bit of "brute force"...

X = 0
(0+2)(0+2) = |0+2|?
(2)(2) = 2? NO, 0 is NOT a solution

X = -1
(-1+2)(-1+2) = |-1+2|
(1)(1) = 1? YES, -1 IS a solution

X = -2
(-2+2)(-2+2) = |-2+2|
(0)(0) = 0? YES, -2 IS a solution

Remember that there were probably going to be MORE than 2 solutions, so we have to keep looking...

X = -3
(-3+2)(-3+2) = |-3+2|
(-1)(-1) = 1? YES, -3 IS a solution

At this point, if we take the product of these 3 answers we have (-1)(-2)(-3) = -6. Looking at the 5 answer choices, there CAN'T be any other answers. To go from -6 to -12, (2) would have to be an answer (but we already know that positive numbers DON'T FIT). To go from -6 to +12, (-2) would have to be used twice, which is not a mathematical option. So -6 MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

If solving for -x-2 gives the solutions x=-3 and x=-2, and solving for x+2 gives the solutions x=-2 and x=-1
why does the answer take into account only -3*-2*-1 = -6 when the question asks for "all solutions"?
If it is asking for all solutions why are we not taking -3*-2*-1*-2 = 12 ?

From the discussions and solutions above, I tried to to rationalize this way:

1) x+2>0 so x>-2

i.e x^2+4x+7=x+2+3
results: x=-2 and -1

checking:
-2+2>0
0>0 {null so doesn't apply}

-1+2>0
1>0 {ok}

2) -(x-2)>0 so x<-2
i.e x^2+4x+7=-x-2+3
results x = -2 and -3

checking:
-(-2)+2>0
4>0 {ok}

-(-3)+2>0
5>0 {ok}

Am I on the right page? I am seeking validation for comprehending the solution of -6 versus the incorrect solution of 12.
Do I treat the repeated value as one solution?
I ask this so if in the future this type of question provides 4 distinct solutions (no repeated value), would we multiply all 4 solutions or do a further check to see the solution holds true against the inequalities?

Solution

Given:

• We are given an equation : \(x^2 + 4x + 7 = |x + 2| + 3\)

To find:

• We need to find the product of all the values of x that satisfy the equation
\(x^2 + 4x + 7 = |x + 2| + 3\)

Approach and Working:

Observe that we have |x+2| in our equation and its values will be :

• -(x+2) for x<-2
• (x+2) for x>= -2

Thus, we need to consider the equation for x<-2 and for x>=-2 differently as the equation will change for both the ranges.

For x<-2 :

• The equation will be\(x^2 + 4x + 7 = -x - 2 + 3\)

o \(x^2+5x+6=0\)
o x= -2,-3

For x>=-2 :

• The equation will be \(x^2 + 4x + 7 = x + 2 + 3\)

o \(x^2+3x+2=0\)
o x= -1, -2

Thus, the product of all the values of x is: -1*-2*-3= -6

Hence, the correct answer is option A.

Answer: A

Some test-takers might find it easiest to proceed as follows:
1. Solve the equation with the signs unchanged
2. Solve the equation with the signs changed in the absolute value
In each case, plug any possible solutions back into the original equation to confirm that they are valid.

\(x^2 + 4x + 7 = |x + 2| + 3\)
\(x^2 + 4x + 4 = |x + 2|\)
Here, a solution will be valid as along as the left side is NONNEGATIVE.

Case 1: signs unchanged
\(x^2 + 4x + 4 = x + 2\)
\(x^2 + 3x + 2 = 0\)
\((x+2)(x+1) = 0\)
Possible solutions: x=-2, x=-1.

If we plug \(x=-2\) into \(x^2 + 4x + 4\), we get:
\((-2)^2 + (4)(-2) + 4 = 4 - 8 + 4 = 0\).
Since the result is nonnegative, x=-2 is a valid solution.

If we plug \(x=-1\) into \(x^2 + 4x + 4\), we get:
\((-1)^2 + (4)(-1) + 4 = 1 - 4 + 4 = 1\).
Since the result is nonnegative, x=-1 is a valid solution.

Case 2: signs changed in the absolute value
\(x^2 + 4x + 4 = -x - 2\)
\(x^2 + 5x + 6 = 0\)
\((x+3)(x+2) = 0\)
Possible solutions: x=-3, x=-2.
Since x=-2 was a valid solution in Case 1, only x=-3 must be tested.

If we plug \(x=-3\) into \(x^2 + 4x + 4\), we get:
\((-3)^2 + (4)(-3) + 4 = 9 - 12 + 4 = 1\).
Since the result is nonnegative, x=-3 is a valid solution.

Product of the solutions = (-2)(-1)(-3) = -6.

Hi all,

I approached the question slightly different in the way I tried to solve.

Firstly, when the right-hand side is positive I set my equation to (x+2)(x+2)=(x+2), thus I divided both sides by (x+2) and was left with x+2=1, which gives me x=-1. However, when the RHS is negative I get (x+2)(x+2)= - (x+2); thus when I divide by x+2, I get x+2=-1, which yields x=-3. So, I am missing x=-2.

What is wrong with my approach here? Why can I not divide out (x+2)?

The equations in blue are both valid if x+2=0.
If x+2=0, then division by x+2 is not allowed.
Implication:
In dividing each equation by x+2, you inadvertently assumed in each case that x+2 was nonzero.
For the first blue equation, we could proceed as follows:

Case 1: x+2 = 0
In this case, x=-2.

Case 2: x+2 ≠ 0
Here -- since x+2 is NONZERO -- we can safely divide each side by x+2:
\((x+2)\frac{x+2}{x+2} = \frac{x+2}{x+2}\)
\(x+2 = 1\)
\(x = -1\).

Thus, the first blue equation has two solutions:
x=-2 and x=-1.

The same line of reasoning could be applied to the second blue equation.

ahhhhh I see. Thank you for your help.

My pleasure.
Be forewarned:
If we apply your approach, we must still confirm that the resulting solutions are valid in the original equation.

Example:
(x-3)(x+1) = |x-3|
If we apply your approach to (x-3)(x+1) = x-3, we get the following solutions:
x=3, x=0.
However, x=0 is not a valid solution for (x-3)(x+1) = |x-3|:
(0-3)(0+1) = |0-3|
-3 = 3.
Thus, the equation in blue yields only one valid solution for (x-3)(x+1) = |x-3|:
x=3.

But even if we take X+2 as nagative, then how X=-2 is valid, it will make the solution 0 ? How is -2 a valid value?

Hi kunalkhanna,

To start, were you comfortable 'simplifying' the given equation down into the following equation...?

(X+2)(X+2) = |X+2|

Looking at this, there's NO WAY that X can be positive. Try TESTing any positive number - the two sides will NEVER equal. So there are NO positive answers.

Looking at the answer choices, they're all within a relatively small distance from 0, so the individual values for X are probably all relatively close to 0 as well. Let's find them with a bit of "brute force"...

X = 0
(0+2)(0+2) = |0+2|?
(2)(2) = 2? NO, 4 does not equal 2, so X = 0 is NOT a solution

X = -1
(-1+2)(-1+2) = |-1+2|
(1)(1) = 1? YES, 1 does equal 1, so X = -1 IS a solution

X = -2
(-2+2)(-2+2) = |-2+2|
(0)(0) = 0? YES, 0 does equal 0, so X= -2 IS a solution

Since we're dealing with a Quadratic and an Absolute Value, there are probably going to be MORE than 2 solutions, so we have to keep looking...

X = -3
(-3+2)(-3+2) = |-3+2|
(-1)(-1) = 1? YES, 1 does equal 1, so X= -3 IS a solution

At this point, if we take the product of these 3 answers we have (-1)(-2)(-3) = -6.

Looking at the 5 answer choices, there CAN'T be any other answers. To go from -6 to -12, "+2" would have to be an answer (but we already know that positive numbers DON'T FIT). To go from -6 to +12, (-2) would have to be used twice, which is not a mathematical option. So -6 MUST be the answer.

GMAT assassins aren't born, they're made,
Rich

the eqn can be simplified as: x^2 + 4x + 4 = |x+2|
this means:

|x+2|=(x+2)^2

so now we have 2 cases to solve for:

x+2=(x+2)^2 --> 1 = (x+2) --> x=-1

then we have

x+2=-(x+2)^2 --> 1 = -(x+2) --> 1 = -x - 2 --> -x = 3 --> x=-3

so my algebra gave me 2 solutions: x= -1 and -3

the solution of -2 was from OBSERVING THE equation rather than the algebra.
Can anyone tell me how i can get -2 without OBSERVING?

regards

Hi Mansoor50,

Your algebraic approach is incomplete. By trying to 'divide out' the Absolute Value, you actually removed the '-2' solution. The inclusion of an Absolute Value almost always means that there will be more than one solution to an equation (and it's almost always a positive one and a negative one); here though, we also have to deal with a squared-term (which ALSO implies more than one solution). If you are going to approach the question algebraically, then you have to account for the fact that (X+2) could be a positive value OR a negative value. In simple terms, you only did half of the work.

GMAT assassins aren't born, they're made,
Rich

The secret here is realising that we are calculating based on the UNIQUE solutions.

If we don't then we end up with 4 possible solutions
x=-2
x= -1
x=-3
x= -2

When we multiply this together (-2)*(-1)*(-3)*(-2) = 12

This is a punishment answer (E).

Easy to get caught by this trap.

Isn't the question unfair in punishing when it doesn't say 'distinct' ?

Hi altairahmad,

In these sorts of Algebraic equations (that ask for the total number of solutions), the word 'distinct' is implied.

For example, how many solutions does the following equation have:

X^2 + 2X + 1 = 0

Does it have one solution or two solutions? When you factor this down, you'll see...

(X+1)(X+1) = 0
X = -1

Even if you write down "-1" twice, there's still just one solution (re: -1); the same logic applies to this prompt (and you're not allowed to 'count' the -2 solution twice).

GMAT assassins aren't born, they're made,
Rich