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

May 23, 2015

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

expanding ${\left(x + 1\right)}^{2} = {\left(x\right)}^{2} + 2. x .1 + {\left(1\right)}^{2} = {x}^{2} + 2 x + 1$

$4 {\left(x + 1\right)}^{2} = 4 \left({x}^{2} + 2 x + 1\right) = 64$
$4 {x}^{2} + 8 x + 4 = 64$

$4 {x}^{2} + 8 x - 60 = 0$

dividing the expression by 4:
${x}^{2} + 2 x - 15 = 0$

Factorising by splitting middle term:
In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot - \left(15\right) = - 15$
And,
${N}_{1} + {N}_{2} = b = 2$
After trying out a few numbers we get ${N}_{1} = 5$ and ${N}_{2} = - 3$
${x}^{2} + 2 x - 15 = {x}^{2} + 5 x - 3 x - 15$
$= x \left(x + 5\right) - 3 \left(x + 5\right)$
$= \left(x + 5\right) \left(x - 3\right)$

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