Mister Exam
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How to use it?
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Identical expressions
- 4z^ two -2 zero z+ twenty-six =0
- 4z squared minus 20z plus 26 equally 0
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Similar expressions
- 4z^2+20z+26=0
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4z^2-20z+26=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:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 4$$
$$b = -20$$
$$c = 26$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$z_{1} = \frac{5}{2} + \frac{i}{2}$$
$$z_{2} = \frac{5}{2} - \frac{i}{2}$$
a*z^2 + b*z + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 4$$
$$b = -20$$
$$c = 26$$
, then
D = b^2 - 4 * a * c =
(-20)^2 - 4 * (4) * (26) = -16
Because D<0, then the equation
has no real roots,
but complex roots is exists.
z1 = (-b + sqrt(D)) / (2*a)
z2 = (-b - sqrt(D)) / (2*a)
or
$$z_{1} = \frac{5}{2} + \frac{i}{2}$$
$$z_{2} = \frac{5}{2} - \frac{i}{2}$$
Vieta's Theorem
rewrite the equation
$$\left(4 z^{2} - 20 z\right) + 26 = 0$$
of
$$a z^{2} + b z + c = 0$$
as reduced quadratic equation
$$z^{2} + \frac{b z}{a} + \frac{c}{a} = 0$$
$$z^{2} - 5 z + \frac{13}{2} = 0$$
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -5$$
$$q = \frac{c}{a}$$
$$q = \frac{13}{2}$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 5$$
$$z_{1} z_{2} = \frac{13}{2}$$
$$\left(4 z^{2} - 20 z\right) + 26 = 0$$
of
$$a z^{2} + b z + c = 0$$
as reduced quadratic equation
$$z^{2} + \frac{b z}{a} + \frac{c}{a} = 0$$
$$z^{2} - 5 z + \frac{13}{2} = 0$$
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -5$$
$$q = \frac{c}{a}$$
$$q = \frac{13}{2}$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 5$$
$$z_{1} z_{2} = \frac{13}{2}$$
Rapid solution
[src]
5 I
z1 = - - -
2 2
$$z_{1} = \frac{5}{2} - \frac{i}{2}$$
5 I
z2 = - + -
2 2
$$z_{2} = \frac{5}{2} + \frac{i}{2}$$
z2 = 5/2 + i/2
Sum and product of roots
[src]
sum
5 I 5 I - - - + - + - 2 2 2 2
$$\left(\frac{5}{2} - \frac{i}{2}\right) + \left(\frac{5}{2} + \frac{i}{2}\right)$$
=
5
$$5$$
product
/5 I\ /5 I\ |- - -|*|- + -| \2 2/ \2 2/
$$\left(\frac{5}{2} - \frac{i}{2}\right) \left(\frac{5}{2} + \frac{i}{2}\right)$$
=
13/2
$$\frac{13}{2}$$
13/2