
The least integral value of $ x $ for which $ {x^2} - 9x + 18 < 0 $ is ________.
Answer
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Hint: In order to this question, to find the least integral of $ x $ for which $ {x^2} - 9x + 18 < 0 $ , we will put the value of $ x $ as $ 1,2,3,.... $ and so on to check whether which is the least integral. We will also find the least integer by finding the range of the given expression.
Complete step-by-step answer:
Given that-
$ {x^2} - 9x + 18 < 0 $
Now, we have to check whether the least integral value of $ x $ will satisfy the given equation.
$ * $ Firstly, we will put $ x = 1 $ :
$ \because {x^2} - 9x + 18 < 0 $
$
\therefore {x^2} - 9x + 18 \\
= {1^2} - 9 \times 1 + 18 \\
= 10 \;
$
And, 10 is not less than 0.
Hence, 1 will not be the least integral of $ x $ .
$ * $ Now, we will put $ x = 2 $ :
$
\therefore {x^2} - 9x + 18 \\
= {2^2} - 9 \times 2 + 18 \\
= 4 - 18 + 18 \\
= 4 \;
$
And, 4 is not less than 0.
Hence, 2 will not be the least integral of $ x $ .
$ * $ Now, we will put $ x = 3 $ :
$
\therefore {x^2} - 9x + 18 \\
= {3^2} - 9 \times 3 + 18 \\
= 9 - 27 + 18 \\
= 0 \;
$
And, 0 is not less than 0, both are equal.
Hence, 3 will not be the least integral of $ x $ .
$ * $ Similarly, we will put $ x = 4 $ :
$
\therefore {x^2} - 9x + 18 \\
= {4^2} - 9 \times 4 + 18 \\
= 16 - 36 + 18 \\
= - 2 \;
$
And, $ - 2 $ is less than 0, or $ - 2 < 0 $ .
Hence, 4 is the least integral of $ x $ for which $ {x^2} - 9x + 18 < 0 $ .
So, the correct answer is “4”.
Note: We can also find the least integral of $ x $ for which $ {x^2} - 9x + 18 < 0 $ by finding the range of the given expression. And to find the range, we have to find the factors first:
$
\therefore {x^2} - 9x + 18 \\
= {x^2} - 6x - 3x + 18 \\
= x(x - 6) - 3(x - 6) \\
= (x - 6)(x - 3) \;
$
So, $ x = 6\,\,\,\,or\,\,\,\,x = 3\, $
Therefore, the range of the given expression is $ \{ 3,6\} $ .
And, the integral of $ x $ for $ {x^2} - 9x + 18 < 0 $ lies between 3 and 6 i.e..4 and 5.
Hence, the least integral is 4.
Complete step-by-step answer:
Given that-
$ {x^2} - 9x + 18 < 0 $
Now, we have to check whether the least integral value of $ x $ will satisfy the given equation.
$ * $ Firstly, we will put $ x = 1 $ :
$ \because {x^2} - 9x + 18 < 0 $
$
\therefore {x^2} - 9x + 18 \\
= {1^2} - 9 \times 1 + 18 \\
= 10 \;
$
And, 10 is not less than 0.
Hence, 1 will not be the least integral of $ x $ .
$ * $ Now, we will put $ x = 2 $ :
$
\therefore {x^2} - 9x + 18 \\
= {2^2} - 9 \times 2 + 18 \\
= 4 - 18 + 18 \\
= 4 \;
$
And, 4 is not less than 0.
Hence, 2 will not be the least integral of $ x $ .
$ * $ Now, we will put $ x = 3 $ :
$
\therefore {x^2} - 9x + 18 \\
= {3^2} - 9 \times 3 + 18 \\
= 9 - 27 + 18 \\
= 0 \;
$
And, 0 is not less than 0, both are equal.
Hence, 3 will not be the least integral of $ x $ .
$ * $ Similarly, we will put $ x = 4 $ :
$
\therefore {x^2} - 9x + 18 \\
= {4^2} - 9 \times 4 + 18 \\
= 16 - 36 + 18 \\
= - 2 \;
$
And, $ - 2 $ is less than 0, or $ - 2 < 0 $ .
Hence, 4 is the least integral of $ x $ for which $ {x^2} - 9x + 18 < 0 $ .
So, the correct answer is “4”.
Note: We can also find the least integral of $ x $ for which $ {x^2} - 9x + 18 < 0 $ by finding the range of the given expression. And to find the range, we have to find the factors first:
$
\therefore {x^2} - 9x + 18 \\
= {x^2} - 6x - 3x + 18 \\
= x(x - 6) - 3(x - 6) \\
= (x - 6)(x - 3) \;
$
So, $ x = 6\,\,\,\,or\,\,\,\,x = 3\, $
Therefore, the range of the given expression is $ \{ 3,6\} $ .
And, the integral of $ x $ for $ {x^2} - 9x + 18 < 0 $ lies between 3 and 6 i.e..4 and 5.
Hence, the least integral is 4.
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