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

1 Answer
Shwetank Mauria · Ryuu
Aug 1, 2017

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

Explanation:

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-2sqrt2)(a+2sqrt2)<0#

Note that

  1. for #a<-2sqrt2#, both #a-2sqrt2# and #a+2sqrt2# are negative and hence #a^2-8>0#
  2. for #a>2sqrt2#, both #a-2sqrt2# and #a+2sqrt2# are positive and hence #a^2-8>0#
  3. But for #-2sqrt2 < a < 2sqrt2#, while #a-2sqrt2# is negative, #a+2sqrt2# 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#.

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