Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation e^x=0
-
Equation x^2+2x-3=0
-
Equation -x-7=x
-
Equation tan(x)=-2
- Express {x} in terms of y where:
- 2*x+2*y=-6
- 12*x-13*y=-14
- 4*x-1*y=-3
- 1*x+5*y=-11
- Graphing y =:
- x^2+4x+5
- Canonical form:
- x^2+4x+5
- Derivative of:
-
x^2+4x+5
-
Identical expressions
- x^ two +4x+ five = zero
- x squared plus 4x plus 5 equally 0
- x to the power of two plus 4x plus five equally zero
- x2+4x+5=0
- x²+4x+5=0
- x to the power of 2+4x+5=0
- x^2+4x+5=O
-
Similar expressions
- x^2-4x+5=0
- x^2+4x-5=0
x^2+4x+5=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
$$a = 1$$
$$b = 4$$
$$c = 5$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$x_{1} = -2 + i$$
$$x_{2} = -2 - i$$
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
$$a = 1$$
$$b = 4$$
$$c = 5$$
, then
D = b^2 - 4 * a * c =
(4)^2 - 4 * (1) * (5) = -4
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
$$x_{1} = -2 + i$$
$$x_{2} = -2 - i$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = 5$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -4$$
$$x_{1} x_{2} = 5$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = 5$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -4$$
$$x_{1} x_{2} = 5$$
Sum and product of roots
[src]
sum
-2 - I + -2 + I
$$\left(-2 - i\right) + \left(-2 + i\right)$$
=
-4
$$-4$$
product
(-2 - I)*(-2 + I)
$$\left(-2 - i\right) \left(-2 + i\right)$$
=
5
$$5$$
5
The graph