Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation cos(x)=√3/2
- Equation x^2-x-56=0
-
Equation 4x^2-12x+9=0
-
Equation 3*x^2-8*x-3=0
- Express {x} in terms of y where:
- -10*x+5*y=10
- -18*x+16*y=4
- -8*x-15*y=-12
- 14*x+3*y=-4
- Canonical form:
- 4xy+3y^2
-
Identical expressions
- 4xy+3y^ two
- 4xy plus 3y squared
- 4xy plus 3y to the power of two
- 4xy+3y2
- 4xy+3y²
- 4xy+3y to the power of 2
-
Similar expressions
- 4xy-3y^2
4xy+3y^2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 3$$
$$b = 4 x$$
$$c = 0$$
, then
The equation has two roots.
or
$$y_{1} = - \frac{2 x}{3} + \frac{2 \sqrt{x^{2}}}{3}$$
$$y_{2} = - \frac{2 x}{3} - \frac{2 \sqrt{x^{2}}}{3}$$
a*y^2 + b*y + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 3$$
$$b = 4 x$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(4*x)^2 - 4 * (3) * (0) = 16*x^2
The equation has two roots.
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
or
$$y_{1} = - \frac{2 x}{3} + \frac{2 \sqrt{x^{2}}}{3}$$
$$y_{2} = - \frac{2 x}{3} - \frac{2 \sqrt{x^{2}}}{3}$$
Vieta's Theorem
rewrite the equation
$$4 x y + 3 y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$\frac{4 x y}{3} + y^{2} = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{4 x}{3}$$
$$q = \frac{c}{a}$$
$$q = 0$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = - \frac{4 x}{3}$$
$$y_{1} y_{2} = 0$$
$$4 x y + 3 y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$\frac{4 x y}{3} + y^{2} = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{4 x}{3}$$
$$q = \frac{c}{a}$$
$$q = 0$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = - \frac{4 x}{3}$$
$$y_{1} y_{2} = 0$$
The graph
Rapid solution
[src]
y1 = 0
$$y_{1} = 0$$
4*re(x) 4*I*im(x)
y2 = - ------- - ---------
3 3
$$y_{2} = - \frac{4 \operatorname{re}{\left(x\right)}}{3} - \frac{4 i \operatorname{im}{\left(x\right)}}{3}$$
y2 = -4*re(x)/3 - 4*i*im(x)/3
Sum and product of roots
[src]
sum
4*re(x) 4*I*im(x)
- ------- - ---------
3 3
$$- \frac{4 \operatorname{re}{\left(x\right)}}{3} - \frac{4 i \operatorname{im}{\left(x\right)}}{3}$$
=
4*re(x) 4*I*im(x)
- ------- - ---------
3 3
$$- \frac{4 \operatorname{re}{\left(x\right)}}{3} - \frac{4 i \operatorname{im}{\left(x\right)}}{3}$$
product
/ 4*re(x) 4*I*im(x)\ 0*|- ------- - ---------| \ 3 3 /
$$0 \left(- \frac{4 \operatorname{re}{\left(x\right)}}{3} - \frac{4 i \operatorname{im}{\left(x\right)}}{3}\right)$$
=
0
$$0$$
0