How do you solve #2x^2-4x-5>0# by algebraically?

1 Answer
Jim G.
Aug 21, 2017

#(-oo,-0.87)uu(2.87,+oo)#

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)#

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