Question

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

Answer 1

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`