Quadratic equation. (Last) easy 10 points!?

x^2-3^2 = 0
(x+3) (x-3) = =
x= 3 or x= -3

x² = 9
x = +/- 3

Answer: x = 3, - 3; (x - 3)(x + 3) are the factors.

thats easiest!!

x^2 = 9
x^2 -9 = 0
x^2 - (3)^2 = 0
using a^2 - b^2 = (a-b) (a+b) we have
(x-3) (x+3) = 0
therefore, x= 3 and x= -3

[-] will be the square root symbol, for I do not know how to make it. Okay. Remember, the quadratic equation is: x=(-b+/-[b^2-4ac])/2a ax^2+bx+c=0 1. x^2-3x-4=0 So a is 1, b is 3, and c is -4. Then plug it in, and simplify! x=(-3 +/- [9-4(-4)]) / 2 x=(-3 +/- [9+16]) / 2 x=(-3 +/- [25]) /2 x=(-3 +/- 5)/2 Now, since you have a +/-, you have to pretend you're in both situations. x=(-3-5)/2=(-8)/2= -4. x=(-3+5)/2=(2)/2= 1. The answer is x = { -4, 1 } 2. x^2-5x+2=0 So, a is 1, b is -5, and c is 2. Then plug it in and simplify. x = (5 +/- [25-8]) / 2 x = (5 +/- [17]) / 2 You cannot simplify any further, for [17] is an irrational number. x = { _5 + [17]_ , _5 - [17]_} .............2.................2.........

x^2 = 9
square root both side

x= +/- 3

That's it Maths rock

x^2 = 9
x^2 - 9 = 0
x^2 + 0x - 9 = 0
x^2 + 3x - 3x - 9 = 0
(x^2 + 3x) - (3x + 9) = 0
x(x + 3) - 3(x + 3) = 0
(x + 3)(x - 3) = 0

x + 3 = 0
x = -3

x - 3 = 0
x = 3

∴ x = ±3

x^2=9
x^2-9=0
(x-3)(x+3)=0

x-3=0
x=3

x+3=0
x=-3

x² = 9
x² – 9 = 0
x² + 0x – 9 = 0
(x + 3)(x – 3) = 0
(x + 3) = 0
x = –3
(x – 3) = 0
x = 3
x = {–3, 3}

(+ or - 3)^2=x^2
x=-3 & x=3

x^2=9 (subtract 9)
x^2-9=0
(x+3)(x-3)=0
x= -3 and x=3