+0

0
63
7

How many integers satisfy the inequality x(3x-4) less than or equal to (6x^2-3x+5)/10

Mar 16, 2020
edited by Guest  Mar 16, 2020
edited by Guest  Mar 16, 2020

#1
0

Present it properly, by editing it, and I will think about helping you.  But you need to be quick,

Mar 16, 2020
#2
+1

x(3x-4) <= (6x^2-3x+5)/10

30x^2 -40x  <= 6x^2 - 3x +5

24x^2-37x-5 <=0   this is a bowl shaped parabola with zeroes (via quadratic formula) at 1.64 and -.101

it is negative in between the zeroes..

.this includes the integers      0 and 1

Mar 16, 2020
edited by Guest  Mar 16, 2020
#4
+1

That is much better. And no the answer is not 6.

$$x(3x-4) \le \frac{(6x^2-3x+5)}{10}\\ 10x(3x-4) \le 6x^2-3x+5\\$$

Now you have to expand the left-hand side

then

take all the terms to the left side so that it is   <=0

then

Factorise the left side.

Now if you let the left side equal y

then y must be less than or equal to 0

The left side, when you put it equal to y, will actually be a concave up parabola.

Find the roots of the parabola.

Any integers between the roots will be solutions to this problem.   :)

Give it a go Mar 16, 2020
#6
-1

....not really here (nor is it the purpose of this forum) to 'give you the answer'   (though this time I did)...but rather to give you methods to solve the problem for YOURSELF.....so at least TRY ! 