Which curve?

Algebra Level 1

( 8 x 3 + 36 x 2 + 54 x + 27 ) ( x 1 ) 2 ( x 3 + 6 x 2 + 12 x + 8 ) ( x + 1 ) 2 < 0 \dfrac{(8x^3 + 36x^2 + 54x + 27){(x-1)}^2}{(x^3 + 6x^2 + 12x + 8){(x+1)}^2} < 0

Find all the possible values of x x .

[ 1 , 3 ) [1 , 3) ( , 4 ] (-\infty , 4] ( 2 , 1.5 ) (-2 , -1.5) [ 4 , ) [4 , \infty)

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2 solutions

Ashish Menon
Apr 16, 2016

Relevant wiki: Wavy Curve Method

( 8 x 3 + 36 x 2 + 54 x + 27 ) ( x 1 ) 2 ( x 3 + 6 x 2 + 12 x + 8 ) ( x + 1 ) 2 < 0 ( 8 x 3 + 27 + 18 x ( 2 x + 3 ) ) ( x 1 ) 2 ( x 3 + 8 + 6 x ( x + 2 ) ) ( x + 1 ) 2 < 0 ( 2 x 3 + 3 3 + ( 3 × 2 x × 3 ) ( 2 x + 3 ) ) ( x 1 ) 2 ( x 3 + 2 3 + ( 3 × x × 2 ) ( x + 2 ) ) ( x + 1 ) 2 < 0 ( 2 x + 3 ) 3 ( x 1 ) 2 ( x + 2 ) 3 ( x + 1 ) 2 < 0 Roots of odd powers = 2 , 1.5 Roots of even powers = 1 , 1 \begin{aligned} \dfrac{(8x^3 + 36x^2 + 54x + 27){(x-1)}^2}{(x^3 + 6x^2 + 12x + 8){(x+1)}^2} & < 0\\ \\ \dfrac{(8x^3 + 27 + 18x(2x + 3)){(x-1)}^2}{(x^3 + 8 + 6x(x + 2)){(x+1)}^2} & < 0\\ \\ \dfrac{({2x}^3 + {3}^3 + (3 × 2x × 3)(2x + 3)){(x-1)}^2}{(x^3 + 2^3 + (3 × x × 2)(x + 2)){(x+1)}^2} & < 0\\ \\ \dfrac{{(2x + 3)}^3{(x-1)}^2}{{(x + 2)}^3{(x+1)}^2} & < 0\\ \\ \text{Roots of odd powers} & = -2 , -1.5\\ \text{Roots of even powers} & = -1 , 1 \end{aligned}

Since the inequality is only < < 0 and not \geq 0 or \leq 0, we dont care about the roots of the denominators.

x ( 2 , 1.5 ) \therefore \boxed{x \in (-2 , -1.5)}

Nice solution! One can also solve this question in 10 sec by eliminating wrong options by trial and error.

A Former Brilliant Member - 5 years, 1 month ago

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What if it was "Less Than or Equal To" Zero? What are we supposed to do about the denominators? Your help will be appreciated.

Shudhanshu Shekhar Mishra - 1 year, 11 months ago

Nice work. An alternative approach is to multiply both sides of the inequality by (denominator) 2 ^2 , so the inequality remains unchanged and you are left with a polynomial.

May I know what app do you use to create this graph?

Pi Han Goh - 5 years, 1 month ago

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Yes, I use GeoGebra :)

Ashish Menon - 5 years, 1 month ago

If it is positive? How to write the answer

V Ramamurthi - 1 year, 1 month ago

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You can solve it the same way by looking at the graph that the author posted.

Pi Han Goh - 1 year, 1 month ago
Harsh Jha
Jan 25, 2017

We can see that 1 is common in 3 options except 1st option, and when we put 1 as x the answer comes to be equal to 0 which is equal to 0. Hence instead of solving this question through wavy curve, we can just put and check for some values of x. Hope this helps you :)

Though this is not the perfect explanantion, this is the best method to obtain answer in less time in competitive exams.

Ashish Menon - 4 years, 4 months ago

It is called trial and error

Siddhesh Shankar - 2 years, 8 months ago

Answer is incorrrect cause -1 should be excluded

Aditya Bk - 1 year, 11 months ago

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