How do you solve #0 = x^2 + 4x + 4 # graphically and algebraically?

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Answer 1 · 2016-08-10T16:47:13

#x = -2#

Algebraically: factorise and then solve each factor = 0.

#(x+2)(x+2) = 0#

#x+2 = 0 rArr x =-2" both give the same answer"#

Graphically:
Draw the graph of #color(blue)(y) = x^2 +4x +4#
This will be a parabola.

To solve #x^2 +4x+4 = color(blue)(0)# , read the point(s) from the graph where the parabola cuts the #x#-axis.

The # x#-axis is the line where # color(blue)(y=0)#

The parabola will intersect (touch) at only point where #x = -2#graph{x^2+4x+4 [-10, 10, -5, 5]}