problems with factoring and completing the square, Please help.?

Question

factor the quadratic expression completely and find the roots of the expression.

1. 135x^2 - 222x + 91

2. 6x^2 - 42x + 72

Completing the square. Find the roots of the quadratic equation.

3. x^2 - 3x = 0

Answer 1

1.
x=(( -b ((+-)) sqrt(b^2-4ac) )) / 2a

x= 0.8667 or 0.7778

2.
= x^2-7x +12 ((the original divided by 6))
=(x-3) (x-4)
x=3 x=4

Answer 2

1. Note that 91 factors to 13*7
Note that 135 factors to 5*3*3*3
No common factors
(15x-13)(9x-7)
x = 13/15, 7/9

2. Note that each terms is divisible by 6
6x^2 - 42x + 72 =
6(x^2 - 7x + 12) =
6(x-3)(x-4)
x = 3, 4

3. to complete the square, what is 1/2 of -3 squared?
1.5 squared = 2.25
Add that to each side of the equation
x^2 - 3x + 2.25 = 2.25
(x - 1.5) = sqrt(2.25)
x - 1.5 = +/- 1.5
x = 0, 3

Answer 3

1)Factors of 91 are 7 and 13
Factors of 135 are 9 and 15
Put them together right and they will come mulitply to the origianal
135x^2 - 222x + 91
(9x-7)(15x-13) = 0
x=7/9 or x =13/15

2)Factor our the common factor of 6.
6(x^2 - 7x + 12)
6(x-4)(x-3) = 0
x=4 or x=3

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

Answer 4

1. (15x- 13)(9x-7)
x=13/15
x=7/9

2. (3x-12)(2x-6)
x=4
x=3

3. x^2 - 3x = 0
x^2 - 3x +9= 0+9
(x-3)^2=9
get the sqrt of both sides
x-3=+-3

x=o x=6

Answer 5

2)6(x^2-7x+12)
6(x-4)(x-3)=0
x=4 x=3 are the roots

3)x^2-3x=0
x(x-3)=0
x=0, x=3 are the roots

1) 135x^2-222x+91
ax^2+bx+c=0
the roots are
x=[-b+ / - sqrt(b^2-4ac)]/2a
a=135 b=-222 c=91