The equation will be 0 only at +- 2/3, so between those, but not including them, the equation will be positive.
Hey guys,
I'm stuck on this tough question ......
Find the values of a which make x^2 - ax + 1 - 2(a^2) always positive for all real x.
I made the expression equal to 0 ..and then since x is real, the discriminant should be greater than or equal to 0 . I manipulated this inequality to get that x lies between -(2/3) and +(2/3), which is actually the answer given , but I am not satisfied since they said that the the expression should be greater than 0 but I set it equal to 0 .
When, in setting f(x)=0, you found the solutions at
then, for "a" real, you have a single solution (graph is tangent to the x-axis) if
You have 2 solutions (graph crosses the x-axis) if
In both of the above cases, f(x) does not lie ENTIRELY above the x-axis.
f(x) lies entirely above the x-axis if
This is because the graph is U-shaped, due to the coefficient of
the highest power of x being positive.
You could also use calculus to examine the minimum value of f(x).