Mister Exam
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Identical expressions
- t^ two +1 zero t+ twenty-one =0
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- t2+10t+21=0
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Similar expressions
- t^2-10t+21=0
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t^2+10t+21=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
$$a = 1$$
$$b = 10$$
$$c = 21$$
, then
Because D > 0, then the equation has two roots.
or
$$t_{1} = -3$$
$$t_{2} = -7$$
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
$$a = 1$$
$$b = 10$$
$$c = 21$$
, then
D = b^2 - 4 * a * c =
(10)^2 - 4 * (1) * (21) = 16
Because D > 0, then the equation has two roots.
t1 = (-b + sqrt(D)) / (2*a)
t2 = (-b - sqrt(D)) / (2*a)
or
$$t_{1} = -3$$
$$t_{2} = -7$$
Vieta's Theorem
it is reduced quadratic equation
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 10$$
$$q = \frac{c}{a}$$
$$q = 21$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = -10$$
$$t_{1} t_{2} = 21$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 10$$
$$q = \frac{c}{a}$$
$$q = 21$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = -10$$
$$t_{1} t_{2} = 21$$
Sum and product of roots
[src]
sum
-7 - 3
$$-7 - 3$$
=
-10
$$-10$$
product
-7*(-3)
$$- -21$$
=
21
$$21$$
21