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Solve the inequation \\\mathrm{\sqrt{x+14}<x+2}

Option: 1

x \in (-2,\infty)


Option: 2

x \in (-14,\infty)


Option: 3

x \in (-14,-5)\cup (2,\infty)


Option: 4

x \in (2,\infty)


Answers (1)

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\sqrt{x+14}<x+2

1. For LHS to be defined, x + 14 ≥ 0 means x ≥ -14

2. When x+2 < 0, then RHS cannot be greater than LHS, so no answer from     this case

3. When x+2 ≥ 0, means when x ≥ -2, 

    In this case we can square both sides

    x + 14 < (x+2)2

    x + 14 < x+ 4x + 4

    x+ 3x - 10 > 0

    (x+5)(x-2) > 0

    x < -5 or x > 2

    Taking intersection with x ≥ -2, which equals x > 2

Now, answer is the intersection of (1) and (3), which is x > 2

 

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Anam Khan

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