We need to use the quadratic formula for this.
-b±√(b2-4ac)
-----------------
2a
Let me explain some of the symbols and letters first.
±- is the plus or minus sign. You will perform both separately when we reach this step.
√-is the square root sign.
The line above 2a is the division line.
The letter a represents the quantity of x2, which is 1 in this case
The letter b represents the quantity of x, which is 5 in this case
The letter c represents the quantity of 1, which is 11 in this case
Let's plug the numbers in.
-5±√(52-4(1)(11))
-----------------
2(1)
Solve:
-5±√(52-4(1)(11))
----------------------- =
2(1)
-5±√((25)-(44))
----------------------- =
2
-5±√((25)-(44))
----------------------- =
2
-5±√(-19)
----------------------- =
2
We cannot have a square root of a negative, so this polynomial is not able to be solved if we restrict ourselves to the real numbers. However, we certainly have two complex roots.
It is a quadratic polynomial.
-5x + 729
7X^3 Third degree polynomial.
2x+5x-24 7x2-24
3(x+y)
None, it involves the square root of a negative number so the roots are imaginary.
-2.5 + 1.6583123951777i-2.5 - 1.6583123951777i
There are none because the discriminant of the given quadratic expression is less than zero.
It is a quadratic polynomial.
-5x + 729
7X^3 Third degree polynomial.
x = -2.5 + 1.6583123951777ix = -2.5 - 1.6583123951777iwhere i is the square root of negative one.
yes, and it is 14x
2x+5x-24 7x2-24
5x(3x+4)
(5x + 2)(x + 1)
2x2+5x+3 = (2x+3)(x+1) when factored