Quadratics Help Please

(x+8)^2 = x+20

What do you do first to solve?

Okay so first you expand the numbers on the left side of the equal sign.

(x + 8)^2 = x + 20
(x + 8)(x + 8) = x + 20
x^2 + 16x + 64 = x + 20

Now collect like terms, and make sure one side of the equal sign is zero.

x^2 + 16x + 64 = x + 20
x^2 + 15x + 44 = 0

Now factor.

x^2 + 15x + 44 = 0
(x + 4)(x + 11) = 0

If any of the values inside one of those brackets equal zero, then the equation is true. What what does x have to be to make any of those zero? We have two answers here.

x = -4 OR -11

If x is -4:

(x + 4)(x + 11) = 0
{-4 + 4)(-4 + 11) = 0
0(7) = 0
0 = 0

If x is -11

(x + 4)(x + 11) = 0
(-11 + 4)(-11 + 11) = 0
(-7)0 = 0
0 = 0

x = -4 or -11

First you deal with the stuff in the brackets by following the rule:

(ax+ b)^2 = ax^2 + 2abx + b^2

So,

(x+8)^2 = x+20

=> x^2 + 2(1)(8)x + 8^2 = x+20

= x^2 + 16x + 64 = x + 20

= x^2 + 16x - x + 64 - 20 = 0

= x^2 + 15x + 44 = 0
factorise. check link in "source" if you don't know how to

after factorising:

(x+ 11) (x+4) = 0

So,

x= -11
or x = -4

(x+8)^2 = x+20
x^2 + 16x + 64 = x + 20
x^2 + 16x + 64 - x - 20 = 0
x^2 + 15x + 44 = 0
(x + 4)(x + 11) = 0
x + 4 = 0, x + 11 = 0
x = - 4, x = - 11

x = -4, -11 answer//

first, expand the left hand side, and move everything from the right hand side to the left hand side to make it equal to zero. then solve.

i.e.

x^2 + 16x +64 = x +20

x^2 +15x +44 = 0

(x+11)(x+4) = 0

therefore
x= -11 or -4

(x+8)^2 is the same as (x+8)(x+8) so solve the brackets first then work out the equation
x^2+16x+64=x+20
x^2+16-x+64-20=0
X^2+15x+44=0
(x+11)(x+4)=0
so X = -11 or -4

(x+8)^2 = x+20
x^2+16x+64 = x+20
x^2+15x+44=0
(x+11)(x+4)=0
x = -11 and -4