How do you solve 2x^2-4x-5>0 by algebraically?
1 Answer
Aug 21, 2017
Explanation:
"obtain the roots of the left side"
2x^2-4x-5rarra=2,b=-4,c=-5
x=(4+-sqrt(16+40))/4
rArrx=1+-1/2sqrt14
rArrx~~ -0.87" or "x~~ 2.87
"since "a>0" then parabola has a minimum turning point"
"y-intercept "=(0,-5)
"graphing the parabola"
graph{2x^2-4x-5 [-10, 10, -5, 5]}
"for "2x^2-4x-5>0
"consider where the graph is above the x-axis"
rArr(-oo,-0.87)uu(2.87,+oo)
