Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation -7x=4
-
Equation x^2-6=0
- Equation x^2-x-90=0
-
Equation x^2+7*x+12=0
- Express {x} in terms of y where:
- -6*x-11*y=-20
- -15*x+7*y=1
- -1*x-20*y=1
- -1*x-15*y=13
- Factor squared:
- x^2-2*x-8
- Derivative of:
- x^2-2*x-8
- Factor polynomial:
- x^2-2*x-8
-
Identical expressions
- x^ two - two *x- eight = zero
- x squared minus 2 multiply by x minus 8 equally 0
- x to the power of two minus two multiply by x minus eight equally zero
- x2-2*x-8=0
- x²-2*x-8=0
- x to the power of 2-2*x-8=0
- x^2-2x-8=0
- x2-2x-8=0
- x^2-2*x-8=O
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Similar expressions
- x^2-2*x+8=0
- x^2+2*x-8=0
x^2-2*x-8=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
$$a = 1$$
$$b = -2$$
$$c = -8$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 4$$
$$x_{2} = -2$$
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
$$a = 1$$
$$b = -2$$
$$c = -8$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (-8) = 36
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 4$$
$$x_{2} = -2$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = -8$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = -8$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = -8$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = -8$$
The graph