help with math question?

Question

Solve the equation by a method of your choice?

1.6x^2+5x+1=0
2.4x^2+15x+9=0
3. x^2-36=0
4.x^2+12+36=0

Answer 1

1.6x^2+5x+1=0
6x^2 +2x+3x+1=0
2x(3x+1) +1(3x+1)=0
(3x+1)(2x+1)=0
x = -1/3 or -1/2

2.4x^2+15x+9=0
(x+3)(4x+3)=0
x = -3 or -3/4

3. x^2-36=0
x^2=36
x= +-9

4.x^2+12x+36=0
(x+6)^2=0
x+6=0
x= -6

Answer 2

There are many ways to take the four equations, but the method of factoring works for all of them. The idea is to factor the left and find values of x that would make one of the terms 0, as anything multiplied by 0 is 0.

1.) 6x^2 + 5x + 1 = (3x + 1)(2x + 1) = 0; x = -1/3, -1/2
2.) 4x^2 + 15x + 9 = (4x + 3)(x + 3) = 0; x = -3/4, -3
3.) x^2 - 36 = (x + 6)(x - 6) = 0; x = -6, 6
4.) x^2 + 12x + 36 = (x + 6)^2 = 0; x = -6

Alternatively, you can use the quadratic equation.