How do you solve $0 = x^{2} + 4 x + 4$ $0 = x^2 + 4x + 4$ graphically and algebraically?

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Answer 1 · 2016-08-10T16:47:13

$x = - 2$ $x = -2$

Algebraically: factorise and then solve each factor = 0.

$(x + 2) (x + 2) = 0$ $(x+2)(x+2) = 0$

$x + 2 = 0 \Rightarrow x = - 2 both give the same answer$ $x+2 = 0 rArr x =-2" both give the same answer"$

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

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

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

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

How do you solve 0 = x^2 + 4x + 4 0=x2+4x+40 = x^2 + 4x + 4 graphically and algebraically?