How do you solve #4(x+1)^2=64#?

1 Answer
May 23, 2015

We know that #color(blue)((a + b)^2 = (a)^2 +2.a.b +(b)^2#

expanding #(x+1)^2= (x)^2 +2.x.1 +(1)^2 = x^2 +2x +1#

#4(x+1)^2= 4(x^2 +2x +1)=64#
#4x^2 +8x +4 = 64#

# 4x^2 + 8x -60 =0#

dividing the expression by 4:
# x^2 + 2x -15 =0#

Factorising by splitting middle term:
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*-(15)= -15#
And,
#N_1 +N_2 = b = 2#
After trying out a few numbers we get #N_1 = 5# and #N_2 =-3#
# x^2 + 2x -15 = x^2 + 5x - 3x -15#
# =x (x+5)-3(x+5)#
# =(x+5)(x-3)#

the solution for the equation is
#x = -5 , x = 3#