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10xy-3+x+2x^2+15y=0 equation
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The solution
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[src]
2 10*x*y - 3 + x + 2*x + 15*y = 0
$$15 y + \left(2 x^{2} + \left(x + \left(10 x y - 3\right)\right)\right) = 0$$
Detail solution
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 2$$
$$b = 10 y + 1$$
$$c = 15 y - 3$$
, then
The equation has two roots.
or
$$x_{1} = - \frac{5 y}{2} + \frac{\sqrt{- 120 y + \left(10 y + 1\right)^{2} + 24}}{4} - \frac{1}{4}$$
$$x_{2} = - \frac{5 y}{2} - \frac{\sqrt{- 120 y + \left(10 y + 1\right)^{2} + 24}}{4} - \frac{1}{4}$$
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 2$$
$$b = 10 y + 1$$
$$c = 15 y - 3$$
, then
D = b^2 - 4 * a * c =
(1 + 10*y)^2 - 4 * (2) * (-3 + 15*y) = 24 + (1 + 10*y)^2 - 120*y
The equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{5 y}{2} + \frac{\sqrt{- 120 y + \left(10 y + 1\right)^{2} + 24}}{4} - \frac{1}{4}$$
$$x_{2} = - \frac{5 y}{2} - \frac{\sqrt{- 120 y + \left(10 y + 1\right)^{2} + 24}}{4} - \frac{1}{4}$$
Vieta's Theorem
rewrite the equation
$$15 y + \left(2 x^{2} + \left(x + \left(10 x y - 3\right)\right)\right) = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + 5 x y + \frac{x}{2} + \frac{15 y}{2} - \frac{3}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 5 y + \frac{1}{2}$$
$$q = \frac{c}{a}$$
$$q = \frac{15 y}{2} - \frac{3}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - 5 y - \frac{1}{2}$$
$$x_{1} x_{2} = \frac{15 y}{2} - \frac{3}{2}$$
$$15 y + \left(2 x^{2} + \left(x + \left(10 x y - 3\right)\right)\right) = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + 5 x y + \frac{x}{2} + \frac{15 y}{2} - \frac{3}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 5 y + \frac{1}{2}$$
$$q = \frac{c}{a}$$
$$q = \frac{15 y}{2} - \frac{3}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - 5 y - \frac{1}{2}$$
$$x_{1} x_{2} = \frac{15 y}{2} - \frac{3}{2}$$
The graph
Rapid solution
[src]
x1 = -3/2
$$x_{1} = - \frac{3}{2}$$
x2 = 1 - 5*re(y) - 5*I*im(y)
$$x_{2} = - 5 \operatorname{re}{\left(y\right)} - 5 i \operatorname{im}{\left(y\right)} + 1$$
x2 = -5*re(y) - 5*i*im(y) + 1
Sum and product of roots
[src]
sum
-3/2 + 1 - 5*re(y) - 5*I*im(y)
$$\left(- 5 \operatorname{re}{\left(y\right)} - 5 i \operatorname{im}{\left(y\right)} + 1\right) - \frac{3}{2}$$
=
-1/2 - 5*re(y) - 5*I*im(y)
$$- 5 \operatorname{re}{\left(y\right)} - 5 i \operatorname{im}{\left(y\right)} - \frac{1}{2}$$
product
-3*(1 - 5*re(y) - 5*I*im(y))
----------------------------
2
$$- \frac{3 \left(- 5 \operatorname{re}{\left(y\right)} - 5 i \operatorname{im}{\left(y\right)} + 1\right)}{2}$$
=
3 15*re(y) 15*I*im(y) - - + -------- + ---------- 2 2 2
$$\frac{15 \operatorname{re}{\left(y\right)}}{2} + \frac{15 i \operatorname{im}{\left(y\right)}}{2} - \frac{3}{2}$$
-3/2 + 15*re(y)/2 + 15*i*im(y)/2