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Find the values of x which make the equation equal to zero. That's easy, you set 2x-4 = 0, and x+6=0 and get the two values of x which solve those. So those are the two points where the function crosses over the x axis. From there, you could see that it's a quadratic with positive leading coefficient (the number in front of x^2 is positive) and know that those functions are are positive for large negative values of x, and positive for large positive values of x, or, you could choose three points, to the left, in the middle, and to the right of the two values of x where the function cross the x axis, and see if it's positive of negative in that area.... You want the points of x which make it less than or equal to zero... so.... they will be the points in between the two critical values of x...
Those two points x on the picture above are the two points you found if you made the thing equal to zero... you can see that in between those, the function is less than or equal to zero.... so you choose all the values of x that are in between those two points, and include the points as well, because it says less than or equal.
14>=2x^(2)+8x-10 2x^(2)+8x-10<=14 2x^(2)+8x-24<=0 Factor out the GCF of 2 from 2x^(2)+8x-24. 2(x^(2)+4x-12)<=0 2(x+6)(x-2)<=0 Divide both sides of the inequality by 2. Dividing 0 by any non-zero number is 0. (x+6)(x-2)<=0 x+6=0 x-2=0 Since this is a 'less than 0' inequality, all intervals that make the expression negative are part of the solution. -6<=x<=2
isn't it -6<=x<2?
\[-6\le x \le2\]
But if you plug 2 back into the equation, x becomes greater than 0, which is incorrect?
If you divide or multiply by a negative number, then the sign will change, the "2" won't change x-2=0 x=2
Yes, I know... but if you plug 2 back into the equation, it becomes greater than 0. But anything less than 2 and anything greater or equal to -6 works...
It includes "2" 2x^(2)+8x-24<=0 2(2)^(2)+8(2)-24<=0 0
Oh, Ok... I was making a silly mistake... :(