square root of x^2+6x+9?

Answer 1

V`(x^2+6x+9)

First: factor the expression. multiply the 1st & 3rd term to get 9. find two numbers that give you 9 when multiplied & 6 (2nd term) when added/subtracted. the numbers are (3 & 3). rewrite the expression with the new middle terms.

x^2 + 3x + 3x + 9

Sec: with 4 terms - group "like" terms & factor both sets of parenthesis.

(x^2 + 3x) + (3x + 9)
x(x + 3) + 3(x + 3)
(x+3)(x+3)

Now, you have V`(x+3)(x+3)

Third: find the square root - you have the same term in the radical sign & the result is...

x+3

Answer 2

x^2 + 6x + 9 is a perfect square polynomial
x^2 + 6x + 9 = (x)^2 + (2*x*3) + (3)^2 = (x + 3)^2

So, square root of x^2 + 6x + 9 is +/-(x + 3)

There are 2 possible answers. They are:
x + 3 and - x -3

Answer 3

x+3

Answer 4

First, rewrite problem. x^2+6x+9
Now, I am going to factor it, then if that isn't what you want, I will still square root it in the way I think you want.
(x+3)^2 is how you factor it, but the to square root that, just take off the ^2, and there is your answer.
Or just square root each part.
Answer is....... x+3

Answer 5

√[x^2+6x+9] = √[(x+3)(x+3)] = x+3.

Answer 6

x^2+6x+9 = (x+3)^2

Square root of (x+3)^2 = +/- (x+3)

So final answer:
x+3
or -x-3

Answer 7

sqrt (x^2+6x+9)
sqrt(x^2 + 3x + 3x + 9)
sqrt(x+3)^2
=±(x+3)

Answer 8

x^2+6x+9 = (x + 3)²

√(x + 3)² = x + 3, -(x + 3)
.
Q.E.D.

Answer 9

x+3, duh