How do you solve #2x^2-7x+3>=0# by graphing?

1 Answer
Feb 22, 2017

Answer:

#x<=1/2" or " x>=3#

Explanation:

Factorise the quadratic and find the roots.

#rArr(2x-1)(x-3)=0#

#rArr"roots are "x=1/2" and " x=3#

#x=0toy=3rArr(0,3)#

#"coefficient of " x^2>0" thus" uu#

With this information we can sketch the function.
graph{2x^2-7x+3 [-10, 10, -5, 5]}

To solve #2x^2+7x-3>=0# Consider the region above and including the x-axis.

Above x-axis to the left of #1/2# and right of 3

#rArrx<=1/2" or " x>=3#