sandy wrote:

How many integers are in the solution set of the inequality \(x^2 - 10 < 0\)?

A. Two

B. Five

C. Six

D. Seven

E. Ten

Practice Questions

Question: 11

Page: 83

Take: \(x^2 - 10 < 0\)

Add 10 to both sides: \(x^2 < 10\)

Let's test some POSITIVE values.

x = 2 satisfies the equation, since \(2^2 < 10\)

x = 3 satisfies the equation, since \(3^2 < 10\)

However, x = 4 does NOT satisfy the equation, since \(4^2 > 10\)

So, the biggest integer value of x is 3.

Let's test some NEGATIVE values.

x = -2 satisfies the equation, since \((-2)^2 < 10\)

x = -3 satisfies the equation, since \((-3)^2 < 10\)

However, x = -4 does NOT satisfy the equation, since \((-4)^2 > 10\)

So, the smallest integer value of x is -3.

When we combine the results, we can see that can be any integer value from -3 to 3 inclusive.

In other words, x can be -3, -2, -1, 0, 1, 2 or 3 (seven possible values)

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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