How Many *Integers* Satisfy

Algebra Level 3

( x 3 ) x x ( x 4 ) 2 ( 17 x ) x ( x 2 + x 1 ) ( x 32 ) < 0 \large \frac {(x-3)^{-\frac {|x|}x}\sqrt{(x-4)^2}(17-x)}{\sqrt{-x}(-x^2+x-1)(|x|-32)} < 0

Find the number of integer x x satisfying the inequality above.


The answer is 31.

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1 solution

Let f ( x ) = p ( x ) q ( x ) f(x) = \dfrac {p(x)}{q(x)} be the LHS of the inequality. Then, we have:

f ( x ) = ( x 3 ) x x ( x 4 ) 2 ( 17 x ) x ( x 2 + x 1 ) ( x 32 ) x is only defined for x 0 f ( x ) is defined for x < 0. = ( x 3 ) 1 x 4 ( 17 x ) x ( x 2 + x 1 ) ( x 32 ) For x < 0 x x = 1 \begin{aligned} f(x) & = \frac {(x-3)^{\color{#D61F06}-\frac {|x|}x}\sqrt{(x-4)^2}(17-x)}{{\color{#3D99F6}\sqrt{-x}}(-x^2+x-1)(|x|-32)} & \small \color{#3D99F6} \sqrt{-x} \text{ is only defined for }x \le 0 \implies f(x) \text{ is defined for }x < 0. \\ & = \frac {(x-3)^{\color{#D61F06}1}|x-4|(17-x)}{\sqrt{-x}(-x^2+x-1)(|x|-32)} & \small \color{#D61F06} \text{For }x < 0 \implies - \frac {|x|}x = 1 \end{aligned}

We note that, for x < 0 x<0 , in the nominator p ( x ) p(x) : { x 3 < 0 x 4 > 0 17 x > 0 \begin{cases} x-3 < 0 \\ |x-4| > 0 \\ 17-x > 0 \end{cases} . Therefore, the nominator p ( x ) < 0 p(x) < 0 . And in the denominator q ( x ) q(x) , for x < 0 x<0 : { x > 0 x 2 + x 1 < 0 \begin{cases} \sqrt{-x} > 0 \\ -x^2+x-1 < 0 \end{cases} . Therefore, f ( x ) < 0 f(x) < 0 only if x 32 < 0 |x|-32<0 or 31 x 1 -31 \le x \le 1 and there are 31 \boxed{31} integral x x satisfying the inequality.

@Ayush Mishra , I have edited your problem. Your original is as below:


If ( x 3 ) x x ( x 4 ) 2 ( 17 x ) x ( x 2 + x 1 ) ( x 32 ) \frac { { \left( x-3 \right) }^{ \frac { -\left| x \right| }{ x } }\sqrt { { \left( x-4 \right) }^{ 2 } } { \left( 17-x \right) } }{ \sqrt { -x } \left( { -x }^{ 2 }+{ x }-{ 1 } \right) \left( \left| x \right| -32 \right) } <0

then no. of integers x satisfying the inequality is


You should put < 0 within LaTex backslash brackets. put x x in LaTex too. Don't use \left| and \right| and \left( and \right) if not necessary. They appear bigger than necessary. Just use | | and ( ). Use \dfrac 12 1 2 \dfrac 12 instead of \frac 12 1 2 \frac 12 for the proper size of fraction.Do not use abbreviation such as "no." just write in full "number". Don't leave the sentence hanging. Instead "the number of integer x x ... is", either "Find the number of integral x x satisfying the inequality above." or "What is the number of integral x x satisfying the inequality above?" Either end with a period (full-stop) or question mark.

Chew-Seong Cheong - 3 years, 3 months ago

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Yes sir I will remember it from now

A Former Brilliant Member - 3 years, 3 months ago

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