Help please someone tell me how to do inequalities and also tell me how you know the sign is < or > . For example, x^2 -11x+24<0, why is the answer 3<x<8 and how do you do it please !!!!!! anwer
Help please someone tell me how to do inequalities and also tell me how you know the sign is < or > . For example, x^2 -11x+24<0, why is the answer 3<x<8 and how do you do it please !!!!!! anwer
You sketch the graph and look for the points where y < 0 or y> 0 etc.
Help please someone tell me how to do inequalities and also tell me how you know the sign is < or > . For example, x^2 -11x+24<0, why is the answer 3<x<8 and how do you do it please !!!!!! anwer
You have to find the critical values.
x2−11x+24 factorises to (x−3)(x−8)
draw that curve and look for the region where y is less than 0.
Help please someone tell me how to do inequalities and also tell me how you know the sign is < or > . For example, x^2 -11x+24<0, why is the answer 3<x<8 and how do you do it please !!!!!! anwer
first, set x2−11x+24=0 and solve for x to find critical values.
when you do this youll get x = 3 or x = 8.
From here you need to work out whether the curve is less than 0 for 3<x<8 or x<3 U x>8. To do this you have two choices;
Sketch it - you can see that the quadratic has a y intercept of 24 so you know three points on the curve are (0,24), (3,0) and (8,0). You also know this quadratic has a U shape so a quick sketch will tell you that x2−11x+24<0 for values 3 < x < 8.
Or you can sub in some values - sub in x = 0, x = 5 and x = 10 for example and youll see that f(0) >0, f(5) < 0 and f(10) > 0 therefore the function is <0 for values 3<x<8
Thanks but how do you look for the region where y is less than 0
All the points of the curve below the x axis. If you look at the graph SeanFM posted, y=0 on the x axis, so at x=3 and x=8, anything below that is where y<0 and anything above is where y>0.