Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 2x^2-8=0
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Equation sin(x)-sqrt(2)/2=0
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Equation x^3-2x^2-9x+18=0
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- Integral of d{x}:
- t^2-t-2
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Identical expressions
- t^ two -t- two = zero
- t squared minus t minus 2 equally 0
- t to the power of two minus t minus two equally zero
- t2-t-2=0
- t²-t-2=0
- t to the power of 2-t-2=0
- t^2-t-2=O
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Similar expressions
- t^2+t-2=0
- t^2-t+2=0
t^2-t-2=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:
$$t_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$t_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -1$$
$$c = -2$$
, then
Because D > 0, then the equation has two roots.
or
$$t_{1} = 2$$
$$t_{2} = -1$$
a*t^2 + b*t + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$t_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$t_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -1$$
$$c = -2$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (1) * (-2) = 9
Because D > 0, then the equation has two roots.
t1 = (-b + sqrt(D)) / (2*a)
t2 = (-b - sqrt(D)) / (2*a)
or
$$t_{1} = 2$$
$$t_{2} = -1$$
Vieta's Theorem
it is reduced quadratic equation
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -1$$
$$q = \frac{c}{a}$$
$$q = -2$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = 1$$
$$t_{1} t_{2} = -2$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -1$$
$$q = \frac{c}{a}$$
$$q = -2$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = 1$$
$$t_{1} t_{2} = -2$$