Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation cos(x+pi/3)=0
-
Equation 2x^2+13-13x=x+1
-
Equation tan(pi*(4x-18)/6)=sqrt(3)/3
- Equation tg^3x-tg^2x-3tgx+3=0
- Express {x} in terms of y where:
- 15*x+15*y=16
- -4*x+18*y=-14
- 8*x+7*y=-13
- 14*x-16*y=14
- Derivative of:
-
x+1
- Graphing y =:
- x+1
- Canonical form:
- x+1
-
Identical expressions
- two x^2+ thirteen - one 3x=x+1
- 2x squared plus 13 minus 13x equally x plus 1
- two x squared plus thirteen minus one 3x equally x plus 1
- 2x2+13-13x=x+1
- 2x²+13-13x=x+1
- 2x to the power of 2+13-13x=x+1
-
Similar expressions
- 2x^2+13-13x=x-1
- 2x^2+13+13x=x+1
- 2x^2-13-13x=x+1
2x^2+13-13x=x+1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$- 13 x + \left(2 x^{2} + 13\right) = x + 1$$
to
$$\left(- 13 x + \left(2 x^{2} + 13\right)\right) + \left(- x - 1\right) = 0$$
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
$$a = 2$$
$$b = -14$$
$$c = 12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 6$$
$$x_{2} = 1$$
left part with negative sign.
The equation is transformed from
$$- 13 x + \left(2 x^{2} + 13\right) = x + 1$$
to
$$\left(- 13 x + \left(2 x^{2} + 13\right)\right) + \left(- x - 1\right) = 0$$
This equation is of the form
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
$$a = 2$$
$$b = -14$$
$$c = 12$$
, then
D = b^2 - 4 * a * c =
(-14)^2 - 4 * (2) * (12) = 100
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 6$$
$$x_{2} = 1$$
Vieta's Theorem
rewrite the equation
$$- 13 x + \left(2 x^{2} + 13\right) = x + 1$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 7 x + 6 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -7$$
$$q = \frac{c}{a}$$
$$q = 6$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 7$$
$$x_{1} x_{2} = 6$$
$$- 13 x + \left(2 x^{2} + 13\right) = x + 1$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 7 x + 6 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -7$$
$$q = \frac{c}{a}$$
$$q = 6$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 7$$
$$x_{1} x_{2} = 6$$
The graph