Mister Exam
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How to use it?
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Equation 2^x=16
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Similar expressions
- 9t²-10t-1=0
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9t²-10t+1=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
$$a = 9$$
$$b = -10$$
$$c = 1$$
, then
Because D > 0, then the equation has two roots.
or
$$t_{1} = 1$$
$$t_{2} = \frac{1}{9}$$
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
$$a = 9$$
$$b = -10$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(-10)^2 - 4 * (9) * (1) = 64
Because D > 0, then the equation has two roots.
t1 = (-b + sqrt(D)) / (2*a)
t2 = (-b - sqrt(D)) / (2*a)
or
$$t_{1} = 1$$
$$t_{2} = \frac{1}{9}$$
Vieta's Theorem
rewrite the equation
$$\left(9 t^{2} - 10 t\right) + 1 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{10 t}{9} + \frac{1}{9} = 0$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{10}{9}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{9}$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{10}{9}$$
$$t_{1} t_{2} = \frac{1}{9}$$
$$\left(9 t^{2} - 10 t\right) + 1 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{10 t}{9} + \frac{1}{9} = 0$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{10}{9}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{9}$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{10}{9}$$
$$t_{1} t_{2} = \frac{1}{9}$$
Sum and product of roots
[src]
sum
1 + 1/9
$$\frac{1}{9} + 1$$
=
10/9
$$\frac{10}{9}$$
product
1/9
$$\frac{1}{9}$$
=
1/9
$$\frac{1}{9}$$
1/9