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Algebra Level 5

− 1 ≤ x 2 + 5 x + a 2 x 2 − 3 x + 2 < 7 \large -1\leq\frac{x^2+5x+a}{2x^2-3x+2}<7 How many integers a a satisfy the inequality above?


The answer is 2.

P C by P C , 20, Vietnam

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1 solution

Case 1 : Find the value of a a that satisfy x 2 + 5 x + a 2 x 2 − 3 x + 2 < 7 \frac{x^2+5x+a}{2x^2-3x+2}<7

x 2 + 5 x + a 2 x 2 − 3 x + 2 < 7 x 2 + 5 x + a 2 x 2 − 3 x + 2 − 7 < 0 − 13 x 2 + 26 x + ( a − 14 ) 2 x 2 − 3 x + 2 < 0 \frac{x^2+5x+a}{2x^2-3x+2}<7 \\ \frac{x^2+5x+a}{2x^2-3x+2}-7<0 \\ \frac{-13x^2+26x+(a-14)}{2x^2-3x+2}<0

Given that, if a x 2 + b x + c ax^2+bx+c has { D < 0 and a < 0 , a x 2 + b x + c always negative D < 0 and a > 0 , a x 2 + b x + c always positive \begin{cases} D \lt 0 & \text{and} & a \lt 0, & ax^2+bx+c & \text{always negative} \\ D \lt 0 & \text{and} & a \gt 0, & ax^2+bx+c & \text{always positive} \end{cases}

Because the denominator has D < 0 D<0 and a > 0 a>0 , 2 x 2 − 3 x + 2 2x^2-3x+2 is always positive. So, value of − 13 x 2 + 26 x + ( a − 14 ) -13x^2+26x+(a-14) must be negative ( D < 0 D<0 ).

D < 0 2 6 2 − 4 ( − 13 ) ( a − 14 ) < 0 676 + 52 a − 728 < 0 a < 1 D<0 \\ 26^2-4(-13)(a-14)<0 \\ 676+52a-728<0 \\ a<1

Case 2 : Find the value of a a that satisfy x 2 + 5 x + a 2 x 2 − 3 x + 2 ≥ − 1 \frac{x^2+5x+a}{2x^2-3x+2} \geq -1

x 2 + 5 x + a 2 x 2 − 3 x + 2 ≥ − 1 x 2 + 5 x + a 2 x 2 − 3 x + 2 + 1 ≥ 0 3 x 2 + 2 x + ( a + 2 ) 2 x 2 − 3 x + 2 ≥ 0 \frac{x^2+5x+a}{2x^2-3x+2} \geq -1 \\ \frac{x^2+5x+a}{2x^2-3x+2}+1 \geq 0 \\ \frac{3x^2+2x+(a+2)}{2x^2-3x+2} \geq 0

Because the denominator value is always positive, 3 x 2 + 2 x + ( a + 2 ) 3x^2+2x+(a+2) is not negative ( 3 x 2 + 2 x + ( a + 2 ) ≥ 0 3x^2+2x+(a+2) \geq 0 , D ≤ 0 D \leq 0 ).

D ≤ 0 4 − 4 ( 3 ) ( a + 2 ) ≤ 0 4 − 12 a − 24 ≤ 0 a ≥ − 5 3 D \leq 0 \\ 4-4(3)(a+2) \leq 0 \\ 4-12a-24 \leq 0 \\ a \geq -\frac{5}{3}

We get the range is − 5 3 ≤ a < 1 -\frac{5}{3} \leq a < 1 . And integer of a a that satisfy the inequalities is − 1 -1 and 0 0 = 2 = \boxed{2} integers.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

If D < 0 D<0 and a < 0 a<0 how a x 2 + b x + c ax^2+bx+c is always negative ?? And also the second equation after that ???

Chirayu Bhardwaj - 5 years, 1 month ago

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Is this as such ,
D < 0 ⟹ b 2 < 4 a c ⟹ b 2 < − 4 c D<0 \implies b^2<4ac \implies b^2<-4c as a a is negative .
But square of any number can't be negative , so c c has to be negative.
But what about b b , it can be negative or positive ?? Please explain. Thank you


Chirayu Bhardwaj - 5 years, 1 month ago

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Sorry for ambiguity. If a < 0 a < 0 , most of the range is negative ( − ∞ < y ≤ maximum value -\infty < y \leq \text{maximum value} ), or the turning point are the maximum value of the function. While for a > 0 a > 0 , most of the range is positive ( minimum value ≤ y < ∞ \text{minimum value} \leq y < \infty ), or the turning point is the minimum value of the function. Because a < 0 a < 0 and D < 0 D < 0 , the function not intercept the x x -axis. So, the function is below the x x -axis, or the function always negative.

Irvan Ary Maulana Nugroho - 5 years ago

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