Mister Exam
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How to use it?
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Equation 9^x-8*3^x-9=0
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Identical expressions
- x^ two - eight *x+ eight = zero
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Similar expressions
- x^2-8*x-8=0
- x^2+8*x+8=0
x^2-8*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
$$a = 1$$
$$b = -8$$
$$c = 8$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2 \sqrt{2} + 4$$
$$x_{2} = 4 - 2 \sqrt{2}$$
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
$$a = 1$$
$$b = -8$$
$$c = 8$$
, then
D = b^2 - 4 * a * c =
(-8)^2 - 4 * (1) * (8) = 32
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2 \sqrt{2} + 4$$
$$x_{2} = 4 - 2 \sqrt{2}$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -8$$
$$q = \frac{c}{a}$$
$$q = 8$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 8$$
$$x_{1} x_{2} = 8$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -8$$
$$q = \frac{c}{a}$$
$$q = 8$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 8$$
$$x_{1} x_{2} = 8$$
Rapid solution
[src]
___ x1 = 4 - 2*\/ 2
$$x_{1} = 4 - 2 \sqrt{2}$$
___ x2 = 4 + 2*\/ 2
$$x_{2} = 2 \sqrt{2} + 4$$
x2 = 2*sqrt(2) + 4
Sum and product of roots
[src]
sum
___ ___ 4 - 2*\/ 2 + 4 + 2*\/ 2
$$\left(4 - 2 \sqrt{2}\right) + \left(2 \sqrt{2} + 4\right)$$
=
8
$$8$$
product
/ ___\ / ___\ \4 - 2*\/ 2 /*\4 + 2*\/ 2 /
$$\left(4 - 2 \sqrt{2}\right) \left(2 \sqrt{2} + 4\right)$$
=
8
$$8$$
8