Question

For what values of a, values of the function y=x2+(a−2)x+0.25 are nonnegative?

Answer 1

Answer:

$a\in (-\infty,1]\cup [3,\infty).$

Step-by-step explanation:

Consider the function $y=x^2+(a-2)x+0.25.$ This function represents the parabola with branches that go in positive y-direction (because the leading coefficient is 1>0).

The disriminant of this quadratic function is

$D=(a-2)^2-4\cdot 1\cdot 0.25=a^2-4a+4-1=a^2-4a+3.$

When the discriminant is ≥0, the quadratic function will take nonnegative values, thus,

$a^2-4a+3\ge 0,\\ \\D=(-4)^2-4\cdot 1\cdot 3=16-12=4,\\ \\a_{1,2}=\dfrac{-(-4)\pm \sqrt{4}}{2\cdot 1}=\dfrac{4\pm 2}{2}=1,\ 3,$

then

$(a-1)(a-3)\ge 0,\\ \\a\in (-\infty,1]\cup [3,\infty).$