Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2-8*x+15=0
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Equation -x^2+4x+3=x^2-x-(1+2x^2)
-
Equation ((57.6/(150-x))^0.15)*((150+x)+10)/171.712=1
-
Equation 3x+3=5x
- Express {x} in terms of y where:
- 20*x+18*y=-9
- -17*x+2*y=9
- -13*x+7*y=-15
- -14*x-8*y=20
- Integral of d{x}:
- x^2+y^2
- Canonical form:
- x^2+y^2
- Plot:
-
x^2+y^2
-
Identical expressions
- x^ two +y^ two = zero
- x squared plus y squared equally 0
- x to the power of two plus y to the power of two equally zero
- x2+y2=0
- x²+y²=0
- x to the power of 2+y to the power of 2=0
- x^2+y^2=O
-
Similar expressions
- x^2+y^2+z^2-3xyz=0
- arctg(x^2+y^2-2x-3)=0
- x^2-y^2=0
x^2+y^2=0 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:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = y^{2}$$
, then
The equation has two roots.
or
$$x_{1} = \sqrt{- y^{2}}$$
$$x_{2} = - \sqrt{- y^{2}}$$
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
$$a = 1$$
$$b = 0$$
$$c = y^{2}$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (y^2) = -4*y^2
The equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \sqrt{- y^{2}}$$
$$x_{2} = - \sqrt{- y^{2}}$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = y^{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 0$$
$$x_{1} x_{2} = y^{2}$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = y^{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 0$$
$$x_{1} x_{2} = y^{2}$$
The graph
Rapid solution
[src]
x1 = -I*re(y) + im(y)
$$x_{1} = - i \operatorname{re}{\left(y\right)} + \operatorname{im}{\left(y\right)}$$
x2 = -im(y) + I*re(y)
$$x_{2} = i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)}$$
x2 = i*re(y) - im(y)
Sum and product of roots
[src]
sum
-I*re(y) + im(y) + -im(y) + I*re(y)
$$\left(- i \operatorname{re}{\left(y\right)} + \operatorname{im}{\left(y\right)}\right) + \left(i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)}\right)$$
=
0
$$0$$
product
(-I*re(y) + im(y))*(-im(y) + I*re(y))
$$\left(- i \operatorname{re}{\left(y\right)} + \operatorname{im}{\left(y\right)}\right) \left(i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)}\right)$$
=
2 -(-im(y) + I*re(y))
$$- \left(i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)}\right)^{2}$$
-(-im(y) + i*re(y))^2