How do you solve and graph #x^2-1>=4x#?

1 Answer
Aug 28, 2017

Answer:

#(-oo,2-sqrt5)uu(2+sqrt5,+oo)#

Explanation:

#color(blue)"graphing the parabola"#

#x^2-4x-1>=0larr" rearranging"#

#"with "a=1,b=-4,c=-1#

#"find zeros using the "color(blue)"quadratic formula"#

#x=(4+-sqrt(16+4))/2=2+-sqrt5#

#rArrx~~ -0.24" or "x~~ 4.24#

#"since "a>0" then minimum turning point "uuu#

#"to find the vertex "color(blue)"complete the square"#

#x^2-4x+4-4-1#

#rArr(x-2)^2-5#

#rArrcolor(magenta)"vertex "=(2,-5)#

#"the solution to the inequality are the values of x "#
#"above the x-axis"#
graph{x^2-4x-1 [-10, 10, -5, 5]}

#rArr(-oo,2-sqrt5)uu(2+sqrt5,+oo)#