Can you help me with these questions?

Question

Solve by factoring.....4x^2+24x+36=0....Thanks. Can you also help me with this one solve by formula....5x^2+9x-2=0...

Answer 1

(i) 4x^2+24x+36=0
Divide across by 4 first:
We get: x^2 + 6x + 9 = 0
Then factorize:

(x+3)(x+3) = 0
So (x+3)^2 = 0
Take the square root of both sides (or at this point you can simply observe that the only value of x which would make this zero is -3)

Taking sqrt, we get: (x+3)=0
Subtract 3 from both sides:
x+3-3 = 0-3
x=-3


(ii) 5x^2+9x-2=0
This is of the general form: ax^2 + bx + c = 0
So the formula is x = -b plus or minus...etc.etc.

So we have a=5, b=9, c=-2:
Then using the formula:

x = -9 +or- sqrt(9^2 - 4*5*-2) all divided by (2*5)
= -9 +or- sqrt(81+40) all divided by 10
= -9 +or- sqrt(121) all divided by 10
= -9 +or- all divided by 10
= (-9-11)/10 OR (-9+11)/10
= -2 or 0.5

Answer 2

4x^2+24x+36=4*(3+x)^2 so -3 is a solution for the first equation

The second equation has solutions -2 and 1/5

Answer 3

Question 1
x² + 6x + 9 = 0
(x + 3).(x + 3) = 0
x = - 3 (twice)

Question 2
x = [- 9 ± √(81 + 40)] / 10
x = [-9 ± 11] / 10
x = 1/5 , x = - 2