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Find the number of integral values of x that satisfy
|x2 + 3x - 1| < 2|x| + 5
    Correct answer is '7'. Can you explain this answer?
    Verified Answer
    Find the number of integral values of x that satisfy|x2 + 3x - 1| <...
    |x2 + 3x - 1| < 2|x| + 5
    We have to find out the boundary points for each modulus and adjust the + or - sign after removing the modulus.
    For the left expression,  x2 + 3x  - 1y = x, the roots are 
    √13 ≈ 3.6
    So, the approx roots are -3.3 and +0.3.
    For the right side of the expression, the boundary point is 0
    Let us denote the same in the number line.

    For R1,
    |x2 + 3x - 1| < 2|x| + 5
    x2 + 3x - 1 < 2x + 5
    x2+ x - 6 < 0
    (x-2) (x +3) < 0
    x ∈ (-3,2)

    Overlapping region is 
    For R2,
    |x2 + 3x - 1| < 2|x| + 5
    -x2 - 3x + 1 < 2x + 5
    -x2 - 5x - 4 < 0
    x2 + 5x + 4 > 0
    (x + 1) (x + 4) > 0
    x ∈ R −(−4,−1)
    Overlapping region is
    For R3,
    |x2 + 3x - 1| < 2|x| + 5
    -x2 - 3x + 1 < - 2x + 5
    - x2 - x - 4 < 0
    - x2 + x + 4 > 0
    This is possible for all values of x. Therefore the overlapping region is
    For R4,
    |x2 + 3x - 1| < 2|x| + 5
    -x2 - 3x + 1 < - 2x + 5
    x2 + 5x - 6 < 0
    (x + 6) (x -1) < 0
    x ∈ (−6, 1)
    Hence, the overall overlapping region is
    Integral values are {-5, -4, -3, -2, -1, 0, 1}
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    Most Upvoted Answer
    Find the number of integral values of x that satisfy|x2 + 3x - 1| <...
    To find the number of integral values of x that satisfy the given expression, we need to determine when the expression inside the absolute value brackets equals 0.

    Setting x^2 + 3x - 1 = 0, we can solve this quadratic equation using factoring or the quadratic formula:

    (x + 1)(x - 1) = 0

    This equation is satisfied when x = -1 or x = 1.

    Therefore, there are 2 integral values of x that satisfy the given expression.
    Free Test
    Community Answer
    Find the number of integral values of x that satisfy|x2 + 3x - 1| <...
    |x2 + 3x - 1| < 2|x| + 5
    We have to find out the boundary points for each modulus and adjust the + or - sign after removing the modulus.
    For the left expression,  x2 + 3x  - 1y = x, the roots are 
    √13 ≈ 3.6
    So, the approx roots are -3.3 and +0.3.
    For the right side of the expression, the boundary point is 0
    Let us denote the same in the number line.

    For R1,
    |x2 + 3x - 1| < 2|x| + 5
    x2 + 3x - 1 < 2x + 5
    x2+ x - 6 < 0
    (x-2) (x +3) < 0
    x ∈ (-3,2)

    Overlapping region is 
    For R2,
    |x2 + 3x - 1| < 2|x| + 5
    -x2 - 3x + 1 < 2x + 5
    -x2 - 5x - 4 < 0
    x2 + 5x + 4 > 0
    (x + 1) (x + 4) > 0
    x ∈ R −(−4,−1)
    Overlapping region is
    For R3,
    |x2 + 3x - 1| < 2|x| + 5
    -x2 - 3x + 1 < - 2x + 5
    - x2 - x - 4 < 0
    - x2 + x + 4 > 0
    This is possible for all values of x. Therefore the overlapping region is
    For R4,
    |x2 + 3x - 1| < 2|x| + 5
    -x2 - 3x + 1 < - 2x + 5
    x2 + 5x - 6 < 0
    (x + 6) (x -1) < 0
    x ∈ (−6, 1)
    Hence, the overall overlapping region is
    Integral values are {-5, -4, -3, -2, -1, 0, 1}
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