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17 May 2024, 10:21
Kudos
Bookmarks
(x+2)²=|x+2|
Squaring both sides
(X+2)⁴=(x+2)²
Let (x+2)²=t
t²-t=0
t(t-1)=0
t=0,t=1
(X+2)²=0,x=-2
(X+2)²=1,x=+-1-2
X=-1 and -3
Product=-6 ans
Posted from my mobile device
Squaring both sides
(X+2)⁴=(x+2)²
Let (x+2)²=t
t²-t=0
t(t-1)=0
t=0,t=1
(X+2)²=0,x=-2
(X+2)²=1,x=+-1-2
X=-1 and -3
Product=-6 ans
Posted from my mobile device
25 May 2024, 06:51
Kudos
Bookmarks
Bunuel
Let, x+2 = a
then (x + 2)^2 = |x + 2| becomes \(a^2 = |a|\) which is possible for
a = ±1 OR a = 0
i.e x+2 = ±1 OR x+2 = 0
if x+2 = 1 then x = -1
if x+2 = -1 then x = -3
if x+2 = 0 then x = -2
Product of all possible values of x = (-1)*(-3)*(-2) = -6
Answer: Option A
22 Jun 2024, 03:36
Kudos
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Bunuel
Keep in mind:
\((x+2)^2 = |x+2|^2\)
Hence we can write the given equation as:\(|x+2|^2 - |x+2| = 0 \)
\(|x+2| * ( |x+2| - 1) = 0\)
So either |x+2| = 0 which means x = -2
or
|x+2| - 1 = 0 which means x is -1 or -3
Product of all solutions = -1 * -2 * -3 = -6
Answer (A)
Video on solving absolute values:
https://youtu.be/oqVfKQBcnrs
