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!
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Linear.
They are parallel.
Simply by using the formula x= -b +/- Underroot of (b2 - 4.a.c) / 2.a 2+x-6=2x+4
Yes it is. The thing that makes it a quadratic equation is that "x squared" in there.
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.
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
They are parallel.
Simply by using the formula x= -b +/- Underroot of (b2 - 4.a.c) / 2.a 2+x-6=2x+4
Yes it is. The thing that makes it a quadratic equation is that "x squared" in there.
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.
They intersect at points (-2/3, 19/9) and (3/2, 5) Solved by combining the two equations together to equal nought and then using the quadratic equation formula to find the values of x and substituting these values into the equations to find the values of y.