105k views
2 votes
For what values of a, values of the function y=x2+(a−2)x+0.25 are nonnegative?

1 Answer

4 votes

Answer:


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

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)=(-(-4)\pm √(4))/(2\cdot 1)=(4\pm 2)/(2)=1,\ 3,

then


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

User MAYANK THAKUR
by
8.4k points

Related questions

1 answer
1 vote
124k views
1 answer
1 vote
75.4k views
asked Nov 4, 2019 19.4k views
Anjsimmo asked Nov 4, 2019
by Anjsimmo
8.0k points
1 answer
4 votes
19.4k views