A Question

Algebra Level 4

Let x x be a positive number such that x 2 + x = 15 \left\lfloor { x }^{ 2 } \right\rfloor +\left\lfloor x \right\rfloor =15 . If the restriction of x x is a x < b a\le x<b , what is ( a + b ) (a+b) ?

Round your answer to 2 decimal places.


The answer is 7.07.

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

Lilian Ciobanu
Dec 13, 2014

3 + 12 ==15 is only because [sqrt(12)]==3

[ [12, 13) ] == 12

a=sqrt(12)==3.464 b=sqrt(13)==3.605

[ [12,13) ] + [ [a,b ) ] == 15

a+b==7.069==7.07

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...