Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation sqrt(x+3)=x-3
- Equation x-y=5
-
Equation -20*(x-13)=-220
-
Equation x-x/12=55/12
- Express {x} in terms of y where:
- 1*x+6*y=-6
- 18*x+19*y=7
- -10*x+13*y=5
- 16*x-15*y=4
- Graphing y =:
- x^2-6*x+10
- Factor squared:
- x^2-6*x+10
-
Identical expressions
- x^ two - six *x+ ten = zero
- x squared minus 6 multiply by x plus 10 equally 0
- x to the power of two minus 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