Mister Exam
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- 3a^ two +a- four = zero
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Similar expressions
- 3a^2+a+4=0
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3a^2+a-4=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:
$$a_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$a_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 3$$
$$b = 1$$
$$c = -4$$
, then
Because D > 0, then the equation has two roots.
or
$$a_{1} = 1$$
$$a_{2} = - \frac{4}{3}$$
a*a^2 + b*a + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$a_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$a_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 3$$
$$b = 1$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (3) * (-4) = 49
Because D > 0, then the equation has two roots.
a1 = (-b + sqrt(D)) / (2*a)
a2 = (-b - sqrt(D)) / (2*a)
or
$$a_{1} = 1$$
$$a_{2} = - \frac{4}{3}$$
Vieta's Theorem
rewrite the equation
$$\left(3 a^{2} + a\right) - 4 = 0$$
of
$$a^{3} + a b + c = 0$$
as reduced quadratic equation
$$a^{2} + b + \frac{c}{a} = 0$$
$$a^{2} + \frac{a}{3} - \frac{4}{3} = 0$$
$$a^{2} + a p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{1}{3}$$
$$q = \frac{c}{a}$$
$$q = - \frac{4}{3}$$
Vieta Formulas
$$a_{1} + a_{2} = - p$$
$$a_{1} a_{2} = q$$
$$a_{1} + a_{2} = - \frac{1}{3}$$
$$a_{1} a_{2} = - \frac{4}{3}$$
$$\left(3 a^{2} + a\right) - 4 = 0$$
of
$$a^{3} + a b + c = 0$$
as reduced quadratic equation
$$a^{2} + b + \frac{c}{a} = 0$$
$$a^{2} + \frac{a}{3} - \frac{4}{3} = 0$$
$$a^{2} + a p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{1}{3}$$
$$q = \frac{c}{a}$$
$$q = - \frac{4}{3}$$
Vieta Formulas
$$a_{1} + a_{2} = - p$$
$$a_{1} a_{2} = q$$
$$a_{1} + a_{2} = - \frac{1}{3}$$
$$a_{1} a_{2} = - \frac{4}{3}$$
Sum and product of roots
[src]
sum
1 - 4/3
$$- \frac{4}{3} + 1$$
=
-1/3
$$- \frac{1}{3}$$
product
-4/3
$$- \frac{4}{3}$$
=
-4/3
$$- \frac{4}{3}$$
-4/3