Stumped on these math problems...HELP PLZ?

4) find the values of x for which the equation is true. you do this by setting each quantity equal to 0:

x-4=0...........................x+3=0

x=4..............................x=-3

x=4 and x=-3 are the roots of the equation. the positive root is x=4

9) first, write the equation in standard form:

2(x^2)-6x+0=0

2(x^2)-6x=0

now factor out 2x:

2x(x-3)=0

(2x)(x-3)=0

now set each quantity equal to 0:

2x=0......................x-3=0

x=0........................x=3

x=0 and x=3 are the roots of the equation.

11) this equation is a quadratic, and it is the difference of two squares. the factoring of it looks like this:

(x-7)(x+7)=0

now set each quantity equal to 0:

x-7=0...................x+7=0

x=7......................x=-7

x=7 and x=-7 are the roots of the equation. the positive root is x=7

another way: you could just: add 49 to both sides:

x^2=49

then take the square root of both sides (but because it's a quadratic, the square root will be ±):

x=±7

12) this equation is a quadratic, and it is the difference of two squares. the factoring of it looks like this:

(x-9)(x+9)=0

now set each quantity equal to 0:

x-9=0...................x+9=0

x=9......................x=-9

x=9 and x=-9 are the roots of the equation. the positive root is x=9

another way: you could just: add 81 to both sides:

x^2=81

then take the square root of both sides (but because it's a quadratic, the square root will be ±):

x=±9

4) x=4

9)2x^2 -6x = 0 ie 2x(x-3)=0 so x=0 or x=3

11) X^2=49 x=sqrt(49) x=7

12)similar to11

4. +4, the negative root would be -3

9. 2x = 6,; x = 3.

2(3)^2 = 6(3)

18 = 18

11. +7, and negative root would be -7

(x-7)(x+7)= 0

12. same technique as above. +9