1)
(x-3)(x-1)
2)
(x+3)(x-1)
3)
(x+4)(x+2)
4)
(x+5)(x-3)
5)
(2x+4)(x-4)
6)
3x(x+2)
7)
(x+4)(x-4)
8)
(2x+4)(2x+8)
9)
(2x+1)(x-3)
10)
have to use quadratic
11)
(x-10)(x-2)
12)
(3x-4)(2x+1)
Do you mean the #12 and severe to fifths to a decimals if you are then it's 12.57
For this case we have that a quadratic equation is of the form:
The roots are given by:
We have the following equation:
We look for the roots:
We have to:
So:
We have two imaginary roots:
Answer:
Write 1 2/3 as an improper fraction...
1 = 3/3. Plus the 2/3 makes 5/3.
To add/subtract fractions, they need the same denominator. (bottom number)
We can change the denominator by multiplying the top and bottom of a fraction by the same number. (it will stay equal, like 1/2 and 2/4)
Let's change 5/3 and 3/4 to a denominator of 12.
5/3 = 20/12 (bottom multiplies by 4, so does the top)
3/4 = 9/12 (bottom multiplies by 3, so does the top)
Now we just subtract the 20 minus 9 twelfths. (the twelfths act as a sort of unit)
If this is just a question to simply this equation, then here's the answer.
-10 + 4n -3t +2t +n
First, add together the n's: 4n+n = 5n
Then, the "t's": 2t - 3t = -t
Since there are no other numbers, you have:
-10 + 5n - t