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Given the equation
[#permalink]
29 Dec 2021, 05:06
3
Given that \(| x^2-2|>1\) and we need to find which of the following could not be a value of x
Let's solve the problem using two methods:
Method 1: Substitution
Let's take each value of x from the answer choice and substitute in the answer choice and see which one satisfies the equation
A. –1. Put x = -1 in \(| x^2-2|>1\) and see if it satisfies the equation. => \(| (-1)^2-2|>1\) => |1-2| > 1 => 1 > 1 which is NOT True.
B. –2. Put x = -2 in \(| x^2-2|>1\) and see if it satisfies the equation. => \(| (-2)^2-2|>1\) => |4-2| > 1 => 4 > 1 which is TRUE
C. –3. Put x = -3 in \(| x^2-2|>1\) and see if it satisfies the equation. => \(| (-3)^2-2|>1\) => |9-2| > 1 => 7 > 1 which is TRUE
D. –4. Put x = -4 in \(| x^2-2|>1\) and see if it satisfies the equation. => \(| (-4)^2-2|>1\) => |16-2| > 1 => 14 > 1 which is TRUE
E. –5. Put x = -5 in \(| x^2-2|>1\) and see if it satisfies the equation. => \(| (-5)^2-2|>1\) => |25-2| > 1 => 23 > 1 which is TRUE
Method 2: Algebra
\(| x^2-2|>1\)
This is of the form |x| > a => x > a or x < -a (Watch this video to learn more about Basics of Absolute Values)
Case 1: \(x^2 - 2 > 1\) => \(x^2\) > 3 => \(x^2 > (\sqrt{3})^2\) => x > \(\sqrt{3}\) or x < -\(\sqrt{3} \) => x > 1.732 or x < -1.732 => x can be -2, -3, -4, -5
Case 2: \(x^2-2 < -1\) \(x^2 < 1\) => -1 < x < 1 (Because \(x^2\) < a => -a < x < a)
So, Answer will be A Hope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems
Re: Given the equation
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03 Apr 2024, 11:37
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