Mister Exam
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Identical expressions
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Similar expressions
- k^2-4*k+3=0
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k^2+4*k+3=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:
$$k_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$k_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 4$$
$$c = 3$$
, then
Because D > 0, then the equation has two roots.
or
$$k_{1} = -1$$
$$k_{2} = -3$$
a*k^2 + b*k + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$k_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$k_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 4$$
$$c = 3$$
, then
D = b^2 - 4 * a * c =
(4)^2 - 4 * (1) * (3) = 4
Because D > 0, then the equation has two roots.
k1 = (-b + sqrt(D)) / (2*a)
k2 = (-b - sqrt(D)) / (2*a)
or
$$k_{1} = -1$$
$$k_{2} = -3$$
Vieta's Theorem
it is reduced quadratic equation
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = 3$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = -4$$
$$k_{1} k_{2} = 3$$
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = 3$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = -4$$
$$k_{1} k_{2} = 3$$
Sum and product of roots
[src]
sum
-3 - 1
$$-3 - 1$$
=
-4
$$-4$$
product
-3*(-1)
$$- -3$$
=
3
$$3$$
3