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18 Oct 2021, 11:45
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viktorija
Hello, everyone. I approached the question in a different manner from what I see posted above, so, in an effort to assist the community, I would like to share. We can rearrange the original equation to make everything easier to work with. Start by isolating the absolute value and then factor.
\(x^2 + 4x + 7 - (3) = |x + 2| + 3 - (3)\)
\(x^2 + 4x + 4 = |x + 2|\)
\((x + 2)^2 = |x + 2|\)
We do not need to go any further. We can appreciate that the only time the value of an expression and its square will be the same is when the square equals 0 or 1. That leaves just three possibilities:
1) the expression equals 0
2) the expression equals -1
3) the expression equals 1
Thus, we need to work with x + 2 on its own.
\(x + 2 = 0\)
\(x = -2\) (solution 1)
\(x + 2 = -1\)
\(x = -3\) (solution 2)
\(x + 2 = 1\)
\(x = -1\) (solution 3)
Thus, x can equal any of -1, -2, and -3. The product is a cinch:
\(-1 * -2 * -3 = -6\)
The answer must be (A). A little knowledge of number properties can go a long way here. Keep it simple. As always, good luck with your studies.
- Andrew
30 Dec 2021, 22:33
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Given that \(x^2 + 4x + 7 = |x + 2| + 3\) and we need to find the product of all the solutions of this equation
To open the Absolute Value we need to consider two cases (Watch this video to learn about the Basics of Absolute Value)
Case 1: x+2 >= 0 or x >= -2
=> |x+2| = x+2 (as x+2 is non-negative)
=> \(x^2 + 4x + 7 = |x + 2| + 3\) = \(x^2 + 4x + 7 = x+2 + 3\)
=> \(x^2 + 4x - x + 7 - 5\) = 0
=> \(x^2 + 3x + 2 \) = 0
=> \(x^2 + x + 2x + 2 \) = 0
=> x(x+1) + 2(x+1) = 0
=> (x+1) (x+2) = 0
=> x = -1 , -2
Both the solutions are in the range x >= -2, so are true
Case 1: x+2 < 0 or x < -2
=> |x+2| = -(x+2) (as x+2 is negative)
=> \(x^2 + 4x + 7 = |x + 2| + 3\) = \(x^2 + 4x + 7 = -(x+2) + 3\)
=> \(x^2 + 4x + x + 7 - 1\) = 0
=> \(x^2 + 5x + 6 \) = 0
=> \(x^2 + 2x + 3x + 6 \) = 0
=> x(x+2) + 3(x+2) = 0
=> (x+2) (x+3) = 0
=> x = -2 , -3
But the range was x < -2, so x = -3 is the solution in this case
=> x = -3, -2, -1
=> Product of the solutions = -3*-2*-1 = -6
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems
To open the Absolute Value we need to consider two cases (Watch this video to learn about the Basics of Absolute Value)
Case 1: x+2 >= 0 or x >= -2
=> |x+2| = x+2 (as x+2 is non-negative)
=> \(x^2 + 4x + 7 = |x + 2| + 3\) = \(x^2 + 4x + 7 = x+2 + 3\)
=> \(x^2 + 4x - x + 7 - 5\) = 0
=> \(x^2 + 3x + 2 \) = 0
=> \(x^2 + x + 2x + 2 \) = 0
=> x(x+1) + 2(x+1) = 0
=> (x+1) (x+2) = 0
=> x = -1 , -2
Both the solutions are in the range x >= -2, so are true
Case 1: x+2 < 0 or x < -2
=> |x+2| = -(x+2) (as x+2 is negative)
=> \(x^2 + 4x + 7 = |x + 2| + 3\) = \(x^2 + 4x + 7 = -(x+2) + 3\)
=> \(x^2 + 4x + x + 7 - 1\) = 0
=> \(x^2 + 5x + 6 \) = 0
=> \(x^2 + 2x + 3x + 6 \) = 0
=> x(x+2) + 3(x+2) = 0
=> (x+2) (x+3) = 0
=> x = -2 , -3
But the range was x < -2, so x = -3 is the solution in this case
=> x = -3, -2, -1
=> Product of the solutions = -3*-2*-1 = -6
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems
26 Mar 2025, 01:24
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There is flaw in this solution {Forgive me if I am wrong}
when you are taking modulus negative then x<-2, so after getting the solution as -3 and -2 we have to eliminate the -2 from here.
{The reason for elimination is consider the equation (x-2) ^2 =0} {here if asked the product of the roots then we will say 4 and not just 2, so we have to count the root 2 twice.}
when you are taking modulus negative then x<-2, so after getting the solution as -3 and -2 we have to eliminate the -2 from here.
{The reason for elimination is consider the equation (x-2) ^2 =0} {here if asked the product of the roots then we will say 4 and not just 2, so we have to count the root 2 twice.}
JeffTargetTestPrep
