Solving Quadratic Equations with Inequalities; Is my answer correst

2k^2-5k+2≤0
Factorise to get (2k-1)≤0 and (k-2)≤0
k=0.5 or k=2
So my answer would be 0.5≤0≥2, is that correct?

If you factorize the equation becomes

(2 k -1 ) ( k - 2) ≤0

But this does NOT mean (2 k -1) ≤ 0 AND (k-2) ≤ 0, because then their product would be positive!


You want either ( + * - or - * + )
( 2 k -1 ≤ 0 AND k-2 ≥ 0 ) or ( 2 k -1 ≥0 AND k-2 ≤ 0)

this means

( k ≤1/2 AND k≥ 2) OR ( k ≥1/2 AND k≤ 2)

The first is an impossibility ( k cannot be both smaller than 1/2 and bigger than 2) , so the second remains: 1/2 ≤ k ≤ 2.

You could also have used the fact that the original equation is that of a parabola with top down, and you know from the factors that the zero's are k = 1/2 and k = 2.
The parabola lies BELOW zero IN BETWEEN these zero's. So the solution can be read off as

1/2 ≤ k ≤ 2

Think of the graph of y=2k^2-5k+2 which is U shaped and crosses the
k axis where 2k-1=0, k=1/2 and k-2=0, k=2
y≤0 for values of x where the graph is below on on the k axis
this is between k=1/2 and k=2 so solution
is 1/2≤k≤2

( 2k - 1 ) ( k - 2 ) ≤ 0
Critical values are k = 1/2 , k = 2

____________ 1/2__________2__________
2k - 1___ -ve___0____+ve________ +ve_____
k - 2____ -ve________-ve____ 0___+ve______
product__+ve___0____-ve___ 0____+ve____

Sol set = { k : 1/2 ≤ k ≤ 2 }

Till second step you are right
For (2k-1) (k-2) < 0 you need 2k-1< 0 or k-2 < 0
If k< 1/2 this condition is satisfied or if k<2 and graeter than 0.5 this condition is satisfied
which means 0.5>k>2

Perfect