Integers

Number Theory Level 3

Find the set of negative integers x x such that ( 1 3 ) x + 4 > ( 1 3 ) x 2 + 3 x + 4 \large \left(\frac{1}{3}\right)^{\sqrt{x+4}} > \left(\frac{1}{3}\right)^{\sqrt{x^2+3x+4}}

If there are X X distinct solutions x 1 , x 2 , ⋯ , x X x_1,x_2,\cdots,x_X , submit X + ∣ ∑ j = 1 X x j ∣ \displaystyle X + \left| \sum_{j=1}^X x_j \right|


The answer is 9.

Vilakshan Gupta by Vilakshan Gupta , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Square-roots may only be taken of non-negative values. Therefore we know immediately that x + 4 ≥ 0 x 2 + 3 x + 4 ≥ 0 x ≥ − 4 ( x + 1 1 2 ) 2 + 1 3 4 ≥ 0 (always true) \begin{array}{ccc} x+4 \geq 0 && x^2 + 3x + 4 \geq 0 \\ x \geq -4 && (x+1\tfrac12)^2 + 1\tfrac34 \geq 0 \\ && \text{(always true)} \end{array} Now for bases b < 1 b < 1 , b x > b y b^x > b^y if and only if x < y x < y . Therefore we must solve x + 4 < x 2 + 3 x + 4 x + 4 < x 2 + 3 x + 4 0 < x 2 + 2 x 0 < x ( x + 2 ) x < − 2 or x > 0. \sqrt{x+4} < \sqrt{x^2 + 3x + 4} \\ x + 4 < x^2 + 3x + 4 \\ 0 < x^2 + 2x \\ 0 < x(x+2) \\ x < -2\ \ \text{or}\ \ x > 0. Thus the solutions are all integers − 4 ≤ x < − 2 -4 \leq x < -2 , that is, x = − 4 , − 3 x = -4, -3 .

We submit the solution 2 + ∣ ( − 4 ) + ( − 3 ) ∣ = 2 + 7 = 9 2 + |(-4) + (-3)| = 2 + 7 = \boxed{9} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

nice solution

Vilakshan Gupta - 3 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...