X^2 - 2X - 13 = 0
X = -b +/- sqrt(b^2-4ac)/2a
a = 1
b = -2
c = -13
X = -(-2) +/- sqrt[(-2)^2 - 4(1)(-13)]/2(1)
X = 2 +/- sqrt(56)/2
2 +/- [2sqrt(14)]/2
Side note: Don't forget to reduce the three twos!
Linear.
They are parallel.
Yes it is. The thing that makes it a quadratic equation is that "x squared" in there.
Simply by using the formula x= -b +/- Underroot of (b2 - 4.a.c) / 2.a 2+x-6=2x+4
3x2 + 2x - 8 = 0 is a quadratic equation.
no only equations with x2 and lower powers can be considered quadratic. those with x3 cannot be considered quadratic, just as x2 cannot be considered linear
It works out that the solutions are: x = 3 and y = 2
Linear.
They are parallel.
If: x+y = 8 and 2x-y = 5 Then by adding the equations together: 3x = 13 or x = 13/3 By substitution into the original equations: x = 13/3 and y = 11/3
Yes it is. The thing that makes it a quadratic equation is that "x squared" in there.
Simply by using the formula x= -b +/- Underroot of (b2 - 4.a.c) / 2.a 2+x-6=2x+4
the system of equations 3x-6y=20 and 2x-4y =3 is?Well its inconsistent.
(2,3)
2x + 5 = 13 2x = 8 2x/2 = 8/2 x=4
3x2 + 2x - 8 = 0 is a quadratic equation.
It is a quadratic equation.