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x^2 + 4x + 7 = |x + 2| + 3
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21 Sep 2021, 08:06
21 Sep 2021, 08:06
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69% (01:48) correct
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\(x^2 + 4x + 7 = |x + 2| + 3\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Product of all the solutions of given equation |
6 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: x^2 + 4x + 7 = |x + 2| + 3
[#permalink]
21 Sep 2021, 10:02
21 Sep 2021, 10:02
3
KarunMendiratta wrote:
\(x^2 + 4x + 7 = |x + 2| + 3\)
Write them as 2 different equations
A) \(x^2 + 4x + 7 = (x + 2) + 3
=> x^2 + 3x + 2 = 0
=> x = -2, -1\)
B) \(x^2 + 4x + 7 = -(x + 2) + 3
=> x^2 + 5x + 6 = 0
=> x = -2, -3\)
Product of -1, -2, and -3 = -6, so B is larger,
Answer is B)
Re: x^2 + 4x + 7 = |x + 2| + 3
[#permalink]
26 Oct 2023, 10:14
26 Oct 2023, 10:14
eskay1981 wrote:
KarunMendiratta wrote:
\(x^2 + 4x + 7 = |x + 2| + 3\)
Write them as 2 different equations
A) \(x^2 + 4x + 7 = (x + 2) + 3
=> x^2 + 3x + 2 = 0
=> x = -2, -1\)
B) \(x^2 + 4x + 7 = -(x + 2) + 3
=> x^2 + 5x + 6 = 0
=> x = -2, -3\)
Product of -1, -2, and -3 = -6, so B is larger,
Answer is B)
Boss, case B is not possible actually. Once you take the 3 to LHS, you'll find that the LHS is (x + 2) ^2 and RHS is |x + 2|. Now, since the square can never be a negative value, the mod cannot take the negative value. Although, x = -3 holds true in the equation. Carcass - did I do something incorrectly?
