Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 1-5*x=-6*x+8
-
Equation x^2+2x-3=0
-
Equation x^2-3*x+3=0
-
Equation 2-3*(2*x+2)=5-4*x
- Express {x} in terms of y where:
- 10*x-2*y=0
- 2*x+1*y=-8
- 2*x+2*y=-6
- 12*x-13*y=-14
- Canonical form:
- x^2+2x+5
- Derivative of:
-
x^2+2x+5
- Integral of d{x}:
-
x^2+2x+5
-
Identical expressions
- x^ two +2x+ five
- x squared plus 2x plus 5
- x to the power of two plus 2x plus five
- x2+2x+5
- x²+2x+5
- x to the power of 2+2x+5
-
Similar expressions
- x^2+2x-5
- x^2-2x+5
x^2+2x+5 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
$$a = 1$$
$$b = 2$$
$$c = 5$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$x_{1} = -1 + 2 i$$
Simplify
$$x_{2} = -1 - 2 i$$
Simplify
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
$$a = 1$$
$$b = 2$$
$$c = 5$$
, then
D = b^2 - 4 * a * c =
(2)^2 - 4 * (1) * (5) = -16
Because D<0, then the equation
has no real roots,
but complex roots is exists.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -1 + 2 i$$
Simplify
$$x_{2} = -1 - 2 i$$
Simplify
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 2$$
$$q = \frac{c}{a}$$
$$q = 5$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -2$$
$$x_{1} x_{2} = 5$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 2$$
$$q = \frac{c}{a}$$
$$q = 5$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -2$$
$$x_{1} x_{2} = 5$$
Sum and product of roots
[src]
sum
0 + -1 - 2*I + -1 + 2*I
$$\left(0 - \left(1 + 2 i\right)\right) - \left(1 - 2 i\right)$$
=
-2
$$-2$$
product
1*(-1 - 2*I)*(-1 + 2*I)
$$1 \left(-1 - 2 i\right) \left(-1 + 2 i\right)$$
=
5
$$5$$
5
The graph