There are different ways to solve a quadratic equation, the main ones that i'm thinking about right now are:
1) factor the equation as a product:
ex: x^2+ 4x + 3 =0
(x+3) (x+1) = 0
x=-3 and x=-1 are the solutions.
To find (x+p) and (x+q) you have to think that (p+q )have to be equal to the number that is multiplied by x, in my example it was 4 (3+1=4), (p times q) have to be equal to the last number of the quadratic equation, the one that is not multiplied by any x, that in my example is 3 (3 x 1= 3)
2) The other way to solve a quadratic function is by using a formula:
given: ax^2 +bx +c=0
x= (-b +/- <span>√(b^2 -</span> 4ac)) / 2a
ex: 3x^2 + 4x -2=0
x= (-4 +/- √16-4(3)(-2)) / 6= (-4 +/- √16+24)/6= (-4 +/- <span>√40) / 6
now there are 2 possibilities: x= (-4+</span><span>√40) /6
and
x= (-4 - </span><span>√40) / 6
I hope the examples were clear enough also if i did't get very nice numbers. Look closely to the sings + and -, they are very important</span>
Adjacent : 53. Is the answer
Answer:
Adding 12 to the circle area is equal to the square area.
Or
s2 = 12 + A
Where
s = side of square
A = area of circle
So
s2 = 12 + 36
s2 = 48
Solve this for s to get the side length
Answer:
it should replace how many numbers
Step-by-step explanation:
we need a pick
