Explained: "For how many integers a is it true..."

Vanessa Hensley

Answered question

2022-01-23

For how many integers a is it true that

a^{2} - 8

is a negative number?

sjkuzy5

Beginner2022-01-24Added 11 answers

Well for

a^{2} - 8

to be a negative number, it tantamounts to solve the inequality

a^{2} - 8 < 0

Now this is nothing but

(a - 2 \sqrt{2}) (a + 2 \sqrt{2}) < 0

1. for

a < - 2 \sqrt{2}

, both

a - 2 \sqrt{2}

and

a + 2 \sqrt{2}

are negative and hence

a^{2} - 8 > 0

2. for

a > 2 \sqrt{2}

, both

a - 2 \sqrt{2}

and

a + 2 \sqrt{2}

are positive and hence

a^{2} - 8 > 0

3. But for

- 2 \sqrt{2} < a < 2 \sqrt{2}

, while

a - 2 \sqrt{2}

is negative,

a + 2 \sqrt{2}

is positive and hence

a^{2} - 8 > 0

Hence possible integer values of a, for which

a^{2} - 8 > 0

are -2,-1,0,1 and 2.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?