How do you find the solution of #x^2-4x>5# algebraically?

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Answer 1 · 2017-06-13T20:06:08

#(-infty,-1)# and #(5,infty)#

Subtract #5# from both sides of the equation.

#x^2-4x-5 > 0#

Graphing this as #y=x^2-4x-5# gives

graph{x^2-4x-5[-3,8,-11,5]}

Notice this graph is positive (i.e., #x^2-4x-5 > 0#) whenever it lies above the #x#-axis.

The key is to find the roots, or zeros, of the quadratic equation.

#x^2-4x-5 = 0#

#(x-5)(x+1) = 0#

#x=5# and #x=-1#

So the solution to the inequality includes the values of #x# on the intervals

#(-infty,-1)# and #(5,infty)#

Graphically, that can be depicted as
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