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I have tried it for multiple times but it still doesn’t work TvT! Please help me eight this question

I have tried it for multiple times but it still doesn’t work TvT! Please help me eight-example-1
User Treasha
by
5.3k points

1 Answer

4 votes

Answer:

a) x = -2

b) 8 integer values

Explanation:

Matthew's solution is correct as far as it goes. However, he failed to take into account the possibility of negative square roots.

His solution might rightly look like ...

3 < 3x² ≤ 75

1 < x² ≤ 25 . . . . . divide by 3

1 < ±x ≤ 5 . . . . . . take the square root

Separating the results:

1 < x ≤ 5

-1 > x ≥ -5 . . . . . . multiply by -1

a)

The value x=-2 shows Matthew's solution is incorrect:

3 < 3(-2)² ≤ 25 ⇒ 3 < 12 ≤ 25 . . . true

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b)

x can have the integer values {-5, -4, -3, -2, 2, 3, 4, 5}, a total of 8 values.

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In the attached, the white space (and dashed line) are the solution set. The shaded areas (and solid line) are excluded from the solution set.

_____

Additional comment

It can be helpful to remember that ...


√(x^2)=|\,x|\\e x

That is, values of x may be negative, but the positive square root will only show you a positive value.

I have tried it for multiple times but it still doesn’t work TvT! Please help me eight-example-1
User Csuwldcat
by
5.4k points