Mister Exam
Other calculators
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How to use it?
- Equation:
- Equation x+y=5
-
Equation x^2+5x+6=0
-
Equation tan(x)=-1/2
-
Equation 1*(x-1)*(x-3)=0
- Express {x} in terms of y where:
- -2*x-19*y=3
- 5*x+4*y=-12
- -18*x-14*y=12
- -9*x+1*y=3
- Graphing y =:
- x^2+6*x+10
- Integral of d{x}:
- x^2+6*x+10
- Factor polynomial:
- x^2+6*x+10
-
Identical expressions
- x^ two + six *x+ ten = zero
- x squared plus 6 multiply by x plus 10 equally 0
- x to the power of two plus six multiply by x plus ten equally zero
- x2+6*x+10=0
- x²+6*x+10=0
- x to the power of 2+6*x+10=0
- x^2+6x+10=0
- x2+6x+10=0
- x^2+6*x+10=O
-
Similar expressions
- x^2+6*x-10=0
- x^2-6*x+10=0
x^2+6*x+10=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
$$a = 1$$
$$b = 6$$
$$c = 10$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$x_{1} = -3 + i$$
$$x_{2} = -3 - i$$
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
$$a = 1$$
$$b = 6$$
$$c = 10$$
, then
D = b^2 - 4 * a * c =
(6)^2 - 4 * (1) * (10) = -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
$$x_{1} = -3 + i$$
$$x_{2} = -3 - i$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 6$$
$$q = \frac{c}{a}$$
$$q = 10$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -6$$
$$x_{1} x_{2} = 10$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 6$$
$$q = \frac{c}{a}$$
$$q = 10$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -6$$
$$x_{1} x_{2} = 10$$
Sum and product of roots
[src]
sum
-3 - I + -3 + I
$$\left(-3 - i\right) + \left(-3 + i\right)$$
=
-6
$$-6$$
product
(-3 - I)*(-3 + I)
$$\left(-3 - i\right) \left(-3 + i\right)$$
=
10
$$10$$
10
The graph