9q^2=6q-1 Can you help me?

Question

Please solve, and explain work! Thanks guys!

Answer 1

ok. subtract 6q and add 1 to the other side:
so, we have:

9q^2 - 6q + 1=0

now, you need to factor:

(3q - 1)(3q - 1)= 0

set 3q - 1 = 0

add 1

3q= 1

divide by 3

q=1/3

Answer 2

this is a quadratic expression,

9q^2 - 6q +1 = 0

Dividing by 9

q^2 - 2/3q + 1/9 = 0

Note 1/3 * 1/3 = 1/9

(q-1/3)(q-1/3) = 0

the roots are 1/3, 1/3

Good luck!

Answer 3

9q^2 = 6q - 1

to solve it, let's first move the (6q - 1) so that we get =0

9q^2 - 6q + 1 = 0

after this, we factorisize it.

(3q - 1) * (3q - 1) = 0

so, we got:
3q = 1
q = 1/3

Answer 4

you can use general form of solution:
q=(-b +_(b^2 - 4ac)^0.5 )/2a
where the equation is:
9q^2-6q+1=0
so ...
a=9 , b= -6 , c= 1 in this case
so ..
q= (6+(36-36)^0.5)/18
= 6/18
= 1/3

Answer 5

9q^2 - 6q + 1 = 0

9q^2 - 3q - 3q + 1
(9q^2-3q) + (-3q+1)
3q(3q-1) + -1(3q-1)
(3q-1)(3q-1)
q = 1/3