quadratic equation always works. In this case, find b^2-4ac.
b^2-4ac=(-28)^2-4(12)(15)=784-720=64
This can be used with quadrat formula, but since it is 8^2, this also means that the equation factors over the integers.
First, quadratic equation: x = (28 +/- sqrt(8^2))/24 = (28 +/- 8)/24 =36/24 or 20/24. Giving 3/2 and 5/6.
Second, use p/q theorem: (qx-p) is a factor, then q divides 12 and p divides 15 (and q can be taken positive).
q=1,2,3,4,6, or 12
p=1,3,5, or 15 and their negatives.
Descarte's Rule of signs tells us that all roots are positive, so p is positive. This still leaves 24 choices that need to be checked!
Let's multiply by 12 first. I did this because the coefficient of x^2 is 12!
12^2x^2 - 28*12x + 15*12 = 0
(12x)^2 - 28(12x) + 180 = 0
So now we can solve everything if we can factor u^2 - 28u + 180 = 0.
This is easier, because q=1 is forced, although we still have 24 factors of 180 to consider. However, there are two such factors that are needed and the add to 28:
(u-10)(u-18) = 0
and our problem is (12x-10)(12x-18) = 0, i.e., after factoring out 2 and 6:
Answers & Comments
Verified answer
12x² - 28x = -15
12x^2 -28x +15 =0
12x^2 -10x -18x +15 =0
2x(6x-5) -3(6x-5) =0
(6x-5)(2x-3) =0
Usually, get all equal to 0 first
12x^2-28x+15=0
quadratic equation always works. In this case, find b^2-4ac.
b^2-4ac=(-28)^2-4(12)(15)=784-720=64
This can be used with quadrat formula, but since it is 8^2, this also means that the equation factors over the integers.
First, quadratic equation: x = (28 +/- sqrt(8^2))/24 = (28 +/- 8)/24 =36/24 or 20/24. Giving 3/2 and 5/6.
Second, use p/q theorem: (qx-p) is a factor, then q divides 12 and p divides 15 (and q can be taken positive).
q=1,2,3,4,6, or 12
p=1,3,5, or 15 and their negatives.
Descarte's Rule of signs tells us that all roots are positive, so p is positive. This still leaves 24 choices that need to be checked!
Let's multiply by 12 first. I did this because the coefficient of x^2 is 12!
12^2x^2 - 28*12x + 15*12 = 0
(12x)^2 - 28(12x) + 180 = 0
So now we can solve everything if we can factor u^2 - 28u + 180 = 0.
This is easier, because q=1 is forced, although we still have 24 factors of 180 to consider. However, there are two such factors that are needed and the add to 28:
(u-10)(u-18) = 0
and our problem is (12x-10)(12x-18) = 0, i.e., after factoring out 2 and 6:
(6x-5)(2x-3) = 0
x=5/6 and x = 3/2 are the solutions.
12x^2 - 28x + 15
(6x - 5)(2x - 3)
x = 5/6 or 3/2
12x^2-28x+15=0
28+/- sq rt -28^2-4(12)(15)/ 24
28+/- sq rt 784-760/24
28+/- sq rt 64/ 24
28+/- 8/ 24
7+/- 8/6
15/6 -1/6
(6x-5)(2x-3)=0
x=5/6 or 3/2 answer
I dnt remember how to do this but u shuld plug numbers in.
i dont think you can factor this, but try the quadratic formula