Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation sin(x/3)=-1/2
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Equation x^3+4*x^2-x-4=0
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Equation 2^2-x-1=x^2-5*x-(-1-x^2)
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Equation (7x-8)-(2x-5)=17
- Express {x} in terms of y where:
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Identical expressions
- two ^ two -x- one =x^ two - five *x-(- one -x^ two)
- 2 squared minus x minus 1 equally x squared minus 5 multiply by x minus ( minus 1 minus x squared )
- two to the power of two minus x minus one equally x to the power of two minus five multiply by x minus ( minus one minus x to the power of two)
- 22-x-1=x2-5*x-(-1-x2)
- 22-x-1=x2-5*x--1-x2
- 2²-x-1=x²-5*x-(-1-x²)
- 2 to the power of 2-x-1=x to the power of 2-5*x-(-1-x to the power of 2)
- 2^2-x-1=x^2-5x-(-1-x^2)
- 22-x-1=x2-5x-(-1-x2)
- 22-x-1=x2-5x--1-x2
- 2^2-x-1=x^2-5x--1-x^2
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Similar expressions
- 2^2-x-1=x^2-5*x+(-1-x^2)
- 2^2-x+1=x^2-5*x-(-1-x^2)
- 2^2-x-1=x^2-5*x-(-1+x^2)
- 2^2+x-1=x^2-5*x-(-1-x^2)
- 2^2-x-1=x^2-5*x-(1-x^2)
- 2^2-x-1=x^2+5*x-(-1-x^2)
2^2-x-1=x^2-5*x-(-1-x^2) equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
2 2 4 - x - 1 = x - 5*x + 1 + x
$$\left(4 - x\right) - 1 = \left(x^{2} + 1\right) + \left(x^{2} - 5 x\right)$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(4 - x\right) - 1 = \left(x^{2} + 1\right) + \left(x^{2} - 5 x\right)$$
to
$$\left(\left(4 - x\right) - 1\right) + \left(\left(- x^{2} - 1\right) + \left(- x^{2} + 5 x\right)\right) = 0$$
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
$$a = -2$$
$$b = 4$$
$$c = 2$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 1 - \sqrt{2}$$
$$x_{2} = 1 + \sqrt{2}$$
left part with negative sign.
The equation is transformed from
$$\left(4 - x\right) - 1 = \left(x^{2} + 1\right) + \left(x^{2} - 5 x\right)$$
to
$$\left(\left(4 - x\right) - 1\right) + \left(\left(- x^{2} - 1\right) + \left(- x^{2} + 5 x\right)\right) = 0$$
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
$$a = -2$$
$$b = 4$$
$$c = 2$$
, then
D = b^2 - 4 * a * c =
(4)^2 - 4 * (-2) * (2) = 32
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 1 - \sqrt{2}$$
$$x_{2} = 1 + \sqrt{2}$$
Vieta's Theorem
rewrite the equation
$$\left(4 - x\right) - 1 = \left(x^{2} + 1\right) + \left(x^{2} - 5 x\right)$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 2 x - 1 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = -1$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = -1$$
$$\left(4 - x\right) - 1 = \left(x^{2} + 1\right) + \left(x^{2} - 5 x\right)$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 2 x - 1 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = -1$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = -1$$
Rapid solution
[src]
___ x1 = 1 - \/ 2
$$x_{1} = 1 - \sqrt{2}$$
___ x2 = 1 + \/ 2
$$x_{2} = 1 + \sqrt{2}$$
x2 = 1 + sqrt(2)
Sum and product of roots
[src]
sum
___ ___ 1 - \/ 2 + 1 + \/ 2
$$\left(1 - \sqrt{2}\right) + \left(1 + \sqrt{2}\right)$$
=
2
$$2$$
product
/ ___\ / ___\ \1 - \/ 2 /*\1 + \/ 2 /
$$\left(1 - \sqrt{2}\right) \left(1 + \sqrt{2}\right)$$
=
-1
$$-1$$
-1
The graph