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

The answer is not 6

Guest Mar 16, 2020

edited by
Guest
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,

Melody 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

ElectricPavlov 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

Melody Mar 16, 2020

#7**0 **

....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 !

ElectricPavlov
Mar 16, 2020