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

Question

Kennedy Adams · Accepted Answer

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#

Eli Assouline · Answer

undefined To solve the quadratic equation x^2 + 4x + 4 = 0 by factoring, you can first recognize that it is a perfect square trinomial, since (x + 2)^2 equals x^2 + 4x + 4. Therefore, you can rewrite the equation as (x + 2)^2 = 0. Then, to solve for x, you take the square root of both sides, giving you x + 2 = 0. Finally, by solving for x, you find that x = -2. Therefore, the solution to the equation x^2 + 4x + 4 = 0 by factoring is x = -2.