How do you solve #x^2+4x+4=0# by factoring?

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Answer 1 · 2015-08-12T06:29:00

The solution is
#color(blue)(x=-2#

# x^2+4x+4=0 #

We can Split the Middle Term of this expression to factorise it and thereby find the solutions:

In this technique, if we have to factorise an expression like #ax^2 + bx + c#, we need to think of 2 numbers such that:

#N_1*N_2 = a*c = 1*4 = 4#
AND
#N_1 +N_2 = b = 4#

After trying out a few numbers we get #N_1 = 2# and #N_2 =2#
#2*2 = 4#, and #2+2= 4#

# x^2+color(blue)(4x)+4= x^2+color(blue)(2x +2x)+4=0 #

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

# color(blue)((x+2)(x+2 )# are the factors of the expression.

Now we equate these factors to zero to find the solutions.

# x+2 = 0 , color(blue)(x=-2#
#x+2 = 0, color(blue)(x=-2#