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How many integers are in the solution set of the inequality
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Updated on: 27 Jan 2019, 02:55
Updated on: 27 Jan 2019, 02:55
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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
A) Two
B) Five
C) Six
D) Seven
E) Ten
Re: How many integers are in the solution set of the inequality
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27 Jan 2019, 03:06
27 Jan 2019, 03:06
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Expert Reply
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The inequality \(x^2 - 10 < 0\) can be written as \(x^2 < 10\)
Now, this inequality basically means (taking the square root) \(x < \pm 3.1\) (3.1 is the square root of ten \(\sqrt {10}\) )
A such, on the number line all numbers in between - 3.1 and + 3.1 are valid numbers
\(-3.1 ----------------0------------- +3.1\)
\(- 3.2\) or \(+ 3.2\) and all the infinite numbers below -3.1 or above +3.1 are impossible because of \(\pm 3.2\) is \(> 10\)
In other words, we are interested in all numbers \(-3.1 < x < +3.1\).
At this point, even in this interval, we do have infinite numbers BUT the stem clearly states we only count the whole numbers
Between -3.1 and +3.1 we do have only seven numbers \({ -3, -2, -1, 0, +1, +2, +3 }\)
D is the answer.
Hope now is clear
Now, this inequality basically means (taking the square root) \(x < \pm 3.1\) (3.1 is the square root of ten \(\sqrt {10}\) )
A such, on the number line all numbers in between - 3.1 and + 3.1 are valid numbers
\(-3.1 ----------------0------------- +3.1\)
\(- 3.2\) or \(+ 3.2\) and all the infinite numbers below -3.1 or above +3.1 are impossible because of \(\pm 3.2\) is \(> 10\)
In other words, we are interested in all numbers \(-3.1 < x < +3.1\).
At this point, even in this interval, we do have infinite numbers BUT the stem clearly states we only count the whole numbers
Between -3.1 and +3.1 we do have only seven numbers \({ -3, -2, -1, 0, +1, +2, +3 }\)
D is the answer.
Hope now is clear
Re: How many integers are in the solution set of the inequality
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27 Jan 2019, 07:06
27 Jan 2019, 07:06
Expert Reply
SKH121 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
A) Two
B) Five
C) Six
D) Seven
E) Ten
\(x^2 – 10 < 0.......x^2<10....|x|<\sqrt{10}\)
As x is an integer, the perfect square just less than 10 is 9, so \(|x|\leq{\sqrt{9}}.......|x|\leq{3}=-3\leq{x}\leq{3}\)
Thus, the integers in the range are -3, -2, -1, 0, 1, 2 and 3, so total 7 values
D
