Mister Exam
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- 3t^ two -28t+ nine = zero
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Similar expressions
- 3t^2+28t+9=0
- 3t^2-28t-9=0
3t^2-28t+9=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
$$a = 3$$
$$b = -28$$
$$c = 9$$
, then
Because D > 0, then the equation has two roots.
or
$$t_{1} = 9$$
$$t_{2} = \frac{1}{3}$$
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
$$a = 3$$
$$b = -28$$
$$c = 9$$
, then
D = b^2 - 4 * a * c =
(-28)^2 - 4 * (3) * (9) = 676
Because D > 0, then the equation has two roots.
t1 = (-b + sqrt(D)) / (2*a)
t2 = (-b - sqrt(D)) / (2*a)
or
$$t_{1} = 9$$
$$t_{2} = \frac{1}{3}$$
Vieta's Theorem
rewrite the equation
$$\left(3 t^{2} - 28 t\right) + 9 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{28 t}{3} + 3 = 0$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{28}{3}$$
$$q = \frac{c}{a}$$
$$q = 3$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{28}{3}$$
$$t_{1} t_{2} = 3$$
$$\left(3 t^{2} - 28 t\right) + 9 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{28 t}{3} + 3 = 0$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{28}{3}$$
$$q = \frac{c}{a}$$
$$q = 3$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{28}{3}$$
$$t_{1} t_{2} = 3$$
Sum and product of roots
[src]
sum
9 + 1/3
$$\frac{1}{3} + 9$$
=
28/3
$$\frac{28}{3}$$
product
9 - 3
$$\frac{9}{3}$$
=
3
$$3$$
3