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a level maths C1

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
Original post by Mr revision
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.



For what values of x is the y value less than 0?
Original post by Mr revision
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.

x211x+24x^2 - 11x + 24 factorises to (x3)(x8)(x-3)(x-8)

draw that curve and look for the region where y is less than 0.
Original post by SeanFM
You sketch the graph and look for the points where y < 0 or y> 0 etc.



For what values of x is the y value less than 0?

Spoiler

Original post by Mr revision
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 x211x+24=0 \displaystyle x^2 - 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 x211x+24<0 \displaystyle x^2 - 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
Thank you all but how do you look for the region where y>0 or y<0
(edited 7 years ago)
Original post by NotNotBatman
You have to find the critical values.

x211x+24x^2 - 11x + 24 factorises to (x3)(x8)(x-3)(x-8)

draw that curve and look for the region where y is less than 0.


Thanks but how do you look for the region where y is less than 0
Original post by Mr revision
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.
(edited 7 years ago)

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