Question

Help me. Please give explanation. Thanks

Answer 1

Answer: x < 3

Step-by-step explanation:

NOTE: $\sqrt{(x+5)^2} = |x + 5|$

Rewrite the equation:

| x + 5 | < | x + 1 |

Remember that absolute value makes the "inside" value positive, so we need to solve for 2 case of x + 5 (inside is positive or negative) and 2 cases of x + 1.

± (x + 5) < ± (x + 1)

Case 1: +  +

x + 5 < x + 1

    5 < 1

  FALSE

Case 2: - -

-(x + 5) < -(x + 1)

 x + 5 > x + 1   divided both sides by -1

      5 > 1

     TRUE so this works when both "insides" are negative

⇒   x + 5 < 0     and      x + 1 < 0

          x < -5     and           x < -1

since it has to satisfy both, x < -5

Case 3: + -

x + 5 < -(x + 1)

x + 5 < -x - 1     distributed -1 on right side

2x + 5 < -1

2x       < -6

 x       < -3

Case 4: - +

-(x + 5) < x + 1

-x - 5 < x + 1     distributed -1 on left side

   -5 < 2x + 1

   -6 < 2x

   -3 < x

     x > -3

This is a contradiction of Case 3 so need to choose a test point.

TEST POINTS (I choose -6, -4, and -2)

 | -6 + 5 | < | -6 + 1 |

 |    -1     | < |    -5   |

          1   <   5

        TRUE so x < -5 is valid

 | -4 + 5 | < | -4 + 1 |

 |     1     | < |    -3   |

          1   <   3

        TRUE so x < -3 is valid

 | -2 + 5 | < | -2 + 1 |

 |     3     | < |    -1   |

          3   <   1

        FALSE so x > 3 is NOT valid

Note that x < -3 includes all the values of x < -5, so x < -3 is the answer.

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This is a faster way to solve it. Thank you @proner26 !

| x + 5 | < | x + 1 |

(| x + 5 |)² < (| x + 1 |)²

x² + 10x + 25 < x² + 2x + 1

       10x + 25 <       2x + 1

        8x + 25 <                1

        8x         <             -24

          x         <              -3

Answer: x < -3