For how many integers $a$ $a$ is it true that $a^{2} - 8$ $a^2-8$ is a negative number?

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Answer 1 · 2017-08-01T04:01:23

Possible integer values of $a$ $a$ , for which $a^{2} - 8 < 0$ $a^2-8<0$ are $- 2, - 1, 0, 1$ $-2,-1,0,1$ and $2$ $2$ .

Well for $a^{2} - 8$ $a^2-8$ to be a negative number, it tantamounts to solve the inequality $a^{2} - 8 < 0$ $a^2-8<0$ .

Now this is nothing but $(a - 2 \sqrt{2}) (a + 2 \sqrt{2}) < 0$ $(a-2sqrt2)(a+2sqrt2)<0$

Note that

for $a < - 2 \sqrt{2}$ $a<-2sqrt2$ , both $a - 2 \sqrt{2}$ $a-2sqrt2$ and $a + 2 \sqrt{2}$ $a+2sqrt2$ are negative and hence $a^{2} - 8 > 0$ $a^2-8>0$
for $a > 2 \sqrt{2}$ $a>2sqrt2$ , both $a - 2 \sqrt{2}$ $a-2sqrt2$ and $a + 2 \sqrt{2}$ $a+2sqrt2$ are positive and hence $a^{2} - 8 > 0$ $a^2-8>0$
But for $- 2 \sqrt{2} < a < 2 \sqrt{2}$ $-2sqrt2 < a < 2sqrt2$ , while $a - 2 \sqrt{2}$ $a-2sqrt2$ is negative, $a + 2 \sqrt{2}$ $a+2sqrt2$ is positive and hence $a^{2} - 8 < 0$ $a^2-8<0$

Hence possible integer values of $a$ $a$ , for which $a^{2} - 8 < 0$ $a^2-8<0$ are $- 2, - 1, 0, 1$ $-2,-1,0,1$ and $2$ $2$ .

For how many integers aa a is it true that a^2-8a2−8a^2-8 is a negative number?