Mister Exam
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How to use it?
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Equation (x^2-4)^2+(x^2-3x-10)^2=0
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Equation 3^x=7
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- 1*x+6*y=-6
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- 2x^2+3y^2
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Identical expressions
- two x^ two +3y^2
- 2x squared plus 3y squared
- two x to the power of two plus 3y squared
- 2x2+3y2
- 2x²+3y²
- 2x to the power of 2+3y to the power of 2
-
Similar expressions
- 2x^2-3y^2
2x^2+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
$$a = 3$$
$$b = 0$$
$$c = 2 x^{2}$$
, then
The equation has two roots.
or
$$y_{1} = \frac{\sqrt{6} \sqrt{- x^{2}}}{3}$$
$$y_{2} = - \frac{\sqrt{6} \sqrt{- x^{2}}}{3}$$
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
$$a = 3$$
$$b = 0$$
$$c = 2 x^{2}$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (3) * (2*x^2) = -24*x^2
The equation has two roots.
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
or
$$y_{1} = \frac{\sqrt{6} \sqrt{- x^{2}}}{3}$$
$$y_{2} = - \frac{\sqrt{6} \sqrt{- x^{2}}}{3}$$
Vieta's Theorem
rewrite the equation
$$2 x^{2} + 3 y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$\frac{2 x^{2}}{3} + y^{2} = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = \frac{2 x^{2}}{3}$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = 0$$
$$y_{1} y_{2} = \frac{2 x^{2}}{3}$$
$$2 x^{2} + 3 y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$\frac{2 x^{2}}{3} + y^{2} = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = \frac{2 x^{2}}{3}$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = 0$$
$$y_{1} y_{2} = \frac{2 x^{2}}{3}$$
The graph
Rapid solution
[src]
___ ___
\/ 6 *im(x) I*\/ 6 *re(x)
y1 = ----------- - -------------
3 3
$$y_{1} = - \frac{\sqrt{6} i \operatorname{re}{\left(x\right)}}{3} + \frac{\sqrt{6} \operatorname{im}{\left(x\right)}}{3}$$
___ ___
\/ 6 *im(x) I*\/ 6 *re(x)
y2 = - ----------- + -------------
3 3
$$y_{2} = \frac{\sqrt{6} i \operatorname{re}{\left(x\right)}}{3} - \frac{\sqrt{6} \operatorname{im}{\left(x\right)}}{3}$$
y2 = sqrt(6)*i*re(x)/3 - sqrt(6)*im(x)/3
Sum and product of roots
[src]
sum
___ ___ ___ ___
\/ 6 *im(x) I*\/ 6 *re(x) \/ 6 *im(x) I*\/ 6 *re(x)
----------- - ------------- + - ----------- + -------------
3 3 3 3
$$\left(- \frac{\sqrt{6} i \operatorname{re}{\left(x\right)}}{3} + \frac{\sqrt{6} \operatorname{im}{\left(x\right)}}{3}\right) + \left(\frac{\sqrt{6} i \operatorname{re}{\left(x\right)}}{3} - \frac{\sqrt{6} \operatorname{im}{\left(x\right)}}{3}\right)$$
=
0
$$0$$
product
/ ___ ___ \ / ___ ___ \ |\/ 6 *im(x) I*\/ 6 *re(x)| | \/ 6 *im(x) I*\/ 6 *re(x)| |----------- - -------------|*|- ----------- + -------------| \ 3 3 / \ 3 3 /
$$\left(- \frac{\sqrt{6} i \operatorname{re}{\left(x\right)}}{3} + \frac{\sqrt{6} \operatorname{im}{\left(x\right)}}{3}\right) \left(\frac{\sqrt{6} i \operatorname{re}{\left(x\right)}}{3} - \frac{\sqrt{6} \operatorname{im}{\left(x\right)}}{3}\right)$$
=
2
-2*(-im(x) + I*re(x))
----------------------
3
$$- \frac{2 \left(i \operatorname{re}{\left(x\right)} - \operatorname{im}{\left(x\right)}\right)^{2}}{3}$$
-2*(-im(x) + i*re(x))^2/3