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

The answer is not 6

 Mar 16, 2020
edited by Guest  Mar 16, 2020
edited by Guest  Mar 16, 2020
 #1
avatar+109509 
0

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

 Mar 16, 2020
 #2
avatar+23566 
+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
 #3
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+1

confusion

 Mar 16, 2020
 #4
avatar+109509 
+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 laugh

 Mar 16, 2020
 #5
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0

what is the answer

Guest Mar 16, 2020
 #6
avatar+109509 
-1

I am not here to give you the answer, I am here to help you learn.

Melody  Mar 16, 2020
 #7
avatar+23566 
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 !   cheeky 

ElectricPavlov  Mar 16, 2020

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