Question

solve the polynomial equation
(x-3)^2=36

Answer 1

x^2-6x= 27

Answer 2

x^2 - 6x -27

Answer 3

x+3 x-9 dont forget medal

Answer 4

(x-3) = +sqrt36 or (x-3) = -sqrt36
x - 3 = 6
x = 9
x - 3 = -6
x = -3

Answer 5

Solve x?

FOIL (x - 3)^2 = (x - 3) (x - 3) = x^2 - 3x - 3x + 9 = 36

Square root the x^2 and the 36

That leaves you with x - 6x + 9 = 6
Combine like terms
-5x + 9 = 6
Subtract 9 on both sides
-5x = -3
Divide by -5 on both sides

Answer 6

That leaves you with -3/-5 which = 3/5!!!

Answer 7

Does that sound correct?

Answer 8

Will the 3/5 check out?
(3/5-3)^2=36
(-12/5)^2=36
144/25=36 Not looking that is a bad answer.

Answer 9

So how do you do it then? :)

Answer 10

Curry above worked it out for you. He may have been to brief but I will redo his work with explanations.
First he expanded the expression enclosed in the parenthesis (x-3)^2
(x-3)(x-3)=x^2-6x+9=36 He then subtracted 9 from both sides getting:
x^2-6x=27 He then subtracted 27 from both sides (leaving 0 on the right)
x^2-6x-27=0 Now it is a quadratic in standard form. It is also factorable
(x-9)(x+3)=0 Now it can be solved as the factors must equal 0 since the product is 0
x-9=0 ......... x must equal 9 x=9 is one root.
x+3=0 ..........x must equal -3, x=-3 the second root. Do you follow this with understanding?