Which values are solutions to the inequality below? Check all that apply.

x² < 144

What are the two answers for this and how do I complete this problem?
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9514 1404 393

Answer:

  -12 < x < 12

Step-by-step explanation:

It helps to think about what this looks like on a graph.

y = x² describes a parabola that has its vertex at the origin and opens upward. You know that 12² = 144, so the value of x² will exceed 144 when the value of x is more than 12. You also know the parabola is symmetric about the y-axis, so the same will be true for values of x less than -12.

The one solution to the inequality is ...

  -12 < x < 12

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You can also go at this slightly differently--a way you might want to use for more complicated quadratic inequalities. Put the inequality in standard form, and factor it.

  x² -144 < 0

  (x +12)(x -12) < 0

Note that these factors are zero for x = -12 and for x = 12. The solution is the set of values of x where the sign of the product is negative.

The sign of the product of two numbers will be negative when one of them is negative and the other is positive. Since the zeros of this product are -12 and +12, the sign of the product will be different in the intervals (-∞, -12), (-12, 12), and (12, ∞). In the left-most of these intervals, both factors are negative, so the product is positive and the interval is not part of the solution set.

For the middle of these intervals, (-12, 12), the sign of (x+12) is positive and the sign of (x-12) is negative. The product will have a negative sign, so this interval is part of the solution set.

For the rightmost interval (12, ∞), both factors have positive signs, so the product is positive and the interval is not in the solution set.

Based on the analysis of the signs of the factors, we realize the solution to the inequality is the interval (-12, 12), or -12 < x < 12.

_____

The wording of your question suggests you are given a list of numbers and asked to identify those that satisfy the inequality. We don't see that list, so we can only tell you the set of numbers the solutions must be in.

Note that -12 or +12 are not solutions to this.

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