How do you solve and write the following in interval notation: #2x^2+x>=10#?

1 Answer
Jul 3, 2016

Answer:

#x <= -5/2#
#x >= 2#

Explanation:

Bring the inequality to standard form:
#f(x) = 2x^2 + x - 10 >= 0#
First, solve the equation f(x) = 0, to find the 2 real roots.
y = (x - 2)(2x + 5)
The 2 real roots are: 2 and #-5/2#
Figure these 2 real roots on a number line. Since a = 2 > 0, the parabola opens upward. Between the 2 real roots, a part of the parabola is below the x, axis, meaning f(x) < 0.
Out side the interval between the 2 real roots, f(x) > 0.
Solution set by half closed intervals: (-inf., -5/2] and [2, inf.).
The 2 end points (2 and -5/2) are included in the solution set.
Graph on number line.

============== -5/2 --------------- 0 ----------- 2 ===============