Can someone help with Quadratic Graphs please?

A Level Quadratic Inequalities - Can Someone Please Help ?

• (2+3x) / (4-x) ≤ 2x I am having difficulty with this one. Can someone please help ? I'd be very grateful. Thanks. Vandomo

• Answer:

  Well, we have : (2+3x) / (4-x) ≤ 2x 1. for : x € (4 , +oo) , 4 ≤ x ---> 4 - x ≤ 0 we obtain : 2+3x ≥ 2x(4-x) 2+3x ≥ 8x - 2x^2 2x^2 - 5x +2 ≥ 0 ∆ = (-5)^2 - 4*2*2 = 9 = 3^2 solutions of trinomial are : x1 = (5 - 3)/4 = 1/2 and x2 = (5 + 3)/4 = 2 -------------> both are < 4 therefore : ==> 2x^2 - x +2 ≥ 0 for any x € (4 , +oo) -------------> part of set of solutions 2. for x € (-oo , 4), 4-x ≥ 0 and we get : 2+3x ≤ 2x(4-x) 2x^2 - 5x +2 ≤ 0 ---> trinomial is the same : it is ≤ 0 BETWEEN THE ROOTS ==> x € [ 1/2 , 2] finally the set of solutions of the initial inequality is : S = [ 1/2 , 2] U (4 , +oo) hope it' ll help !! PS: pls don't forget to give Best Answers, because according to "new rules" :( only the asker may give Best Answers. To me or to anybody else !! ---> it's about keeping people motivated to answer! michael