I will use the quadratic equation here. By inspection the discriminant shows 2 real roots.
X = -b +/- sqrt(b^2 - 4ac)/2a
a = 1
b = - 3
c = - 2
X = -(-3) +/- sqrt[(-3)^2 - 4(1)(-2)]2(1)
X = 3 +/- sqrt(17)/2
X = [3 +/- sqrt(17]/2
exact answer.
No. Y=3X2+1 is a Binomial Equation.
3x2 + 2x - 8 = 0 is a quadratic equation.
X = 5 or -5 3x2 - 42 = 33 3x2 = 75 x2=25
3x2 + 12x = - 1 ie 3x2 + 12x + 1 = 0 has no rational roots. The irrational roots are [-12 +/- sqrt(132)]/6 = -3.915 and -0.085
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
No. Y=3X2+1 is a Binomial Equation.
3x2 + 2x - 8 = 0 is a quadratic equation.
X = 5 or -5 3x2 - 42 = 33 3x2 = 75 x2=25
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
Yes. (Assuming that -3x2 is the best representation of 3x2 that this browser will allow.)
Yes.
3x2 + 12x = - 1 ie 3x2 + 12x + 1 = 0 has no rational roots. The irrational roots are [-12 +/- sqrt(132)]/6 = -3.915 and -0.085
y2=x3+3x2
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
It is: 3x2+6x-11 = 0
x = +/- sqrt(y/3)
2