(s+4)^2=49 (solve equation by factoring)?

Answer 1

(s+4)^2=49
(s+4)^2 - 49 = 0
(s+4)^2 - 7^2 = 0
(s+4+7)(s+4-7) = 0
(s+11)(s-3) = 0

s = {-11, 3}

Answer 2

(s + 4)^2 = 49
First subtract both sides by 49 to have a 0 on the right side.

(s+4)^2 - 49 = 0
We can rewrite this equation as
(s+4)^2 - 7^2 = 0
and use a^2 - b^2 = (a - b)(a + b)
by letting a = (s + 4) and b = 7

so, (s + 4 - 7)(s + 4 + 7) = 0
(s - 3)(s + 11) = 0

Setting both factors to 0
s - 3 = 0
s = 3

s + 11 = 0
s = -11

Therefore the solution is s = 3 and s = -11

Answer 3

To solve this equation by factoring (as opposed to taking teh square root of both sides then isolating x, etc.): First multiply out the left side, then bring the 49 over to the left, and refactor.

s^2 + 8s + 16 = 49
s^2 + 8s + 16 - 49 = 0
s^2 + 8s + 16 - 49 = 0
s^2 + 8s - 33 = 0
(s + 11)(s - 3) = 0
So s = -11 or 3

Answer 4

1st method
(s + 4)^2 = 49
s^2 + 8s + 16 = 49
s^2 + 8s - 33 = 0
(s + 11)(s - 3) = 0
s = 3 or s = -11

or

2nd method
(s + 4)^2 = 49; get the square root on both sides of eq. to get
s + 4 = 7; s = 7 - 4
s = 3

You will know that my answer is correct by subst. 3 or -11 back into the equation!

Answer 5

(s+4)²=49

Expand the left side:
s²+8s+16 = 49

Subtract 49 from both sides:
s²+8s-33 = 0

Factor the trinomial into two binomials:
(s + 11)(s - 3) = 0

Use zero-product property, set each binomial factor equal to zero:
s + 11 = 0 or s - 3 = 0
s = -11 or s = 3

Answer 6

s² + 8s + 16 = 49
s² + 8s - 33 = 0
(s + 11) (s - 3) = 0
s = - 11 , s = 3

Answer 7

hi

(s+4)^2 = 49
(s+4)^2 - 49 = 0
(s+4)^2 - 7^2 = 0
[(s+4)- 7] * [(s+4) + 7] = 0
[ s - 3] * [s+11 ] = 0

then s1 = 3 ; s2 = -11

bye

Answer 8

you dont need to factor, just take the square root of each side, which gives you s+4=7, and s + 4 = -7 then solve for s