9x^2 – 12x + 4 = 0
2007-03-18 06:29:55 · 5 answers · asked by blahblah 1 in Science & Mathematics ➔ Mathematics
First you need to factor it. Start by multiplying 9 x 4 (36). Now what two factors of positive 36 add up to -12? So it would be -6 and -6. Then I use a method called expansion and grouping, so first you expand the polynomial using your two factors, like so:
9x^2 - 6x - 6x + 4 = 0 (See how -6x and -6x make -12x)
Now you group it:
(9x^2 - 6x) + (-6x + 4)
Next you factor each group in parentheses:
3x(3x - 2)-2(3x - 2) The resulting factors are (3x - 2)(3x -2) because you take what's in parentheses (they will always be the same), and it is one factor, and then the two monomials pulled out make the other factor. This one happens to be a perfect square trinomial so the two factors are the same. Now you can solve by setting 3x - 2 equal to 0.
The answer is {2/3 DR} (DR means double root) because there are two factors of (3x -2)
2007-03-18 06:42:34 · answer #1 · answered by purplegrl28 4 · 0⤊ 0⤋
As written it is a quadratic equation not a function so I presume you mean what are the zeros of
f(x) = 9x^2 -12x + 4.
You need to factorise it. Hint: the factors are the same.
2007-03-18 13:35:20 · answer #2 · answered by mathsmanretired 7 · 0⤊ 0⤋
x1,2 = [12 +- sqrt(12^2 - 4*4*9)]/18 = 12/18 = 2/3
2007-03-18 13:36:24 · answer #3 · answered by Amit Y 5 · 0⤊ 0⤋
this function is solved at (3x-2)(3x-2)... hence it intesects the x axis only at x=3/2. which is the zero for the function...
2007-03-18 13:35:34 · answer #4 · answered by Ayush Javeri 2 · 0⤊ 0⤋
a = 9
b= -12
c = 4
Now plug in and solve!!!!
2007-03-18 13:32:32 · answer #5 · answered by J Z 4 · 0⤊ 0⤋