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- Equation:
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Equation 3*x^2-8*x-3=0
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Equation x^3=9
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Equation z^4+z^2+1=0
- Equation z^3=-i
- Express {x} in terms of y where:
- -10*x+5*y=10
- -18*x+16*y=4
- 6*x+9*y=15
- -8*x-15*y=-12
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Identical expressions
- seven , three (x− fifteen)(x+ thirty)= zero
- 7,3(x−15)(x plus 30) equally 0
- seven , three (x− fifteen)(x plus thirty) equally zero
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Similar expressions
- 7,3(x−15)(x-30)=0
7,3(x−15)(x+30)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
73*(x - 15)
-----------*(x + 30) = 0
10
$$\frac{73 \left(x - 15\right)}{10} \left(x + 30\right) = 0$$
Detail solution
Expand the expression in the equation
$$\frac{73 \left(x - 15\right)}{10} \left(x + 30\right) = 0$$
We get the quadratic equation
$$\frac{73 x^{2}}{10} + \frac{219 x}{2} - 3285 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \frac{73}{10}$$
$$b = \frac{219}{2}$$
$$c = -3285$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 15$$
$$x_{2} = -30$$
$$\frac{73 \left(x - 15\right)}{10} \left(x + 30\right) = 0$$
We get the quadratic equation
$$\frac{73 x^{2}}{10} + \frac{219 x}{2} - 3285 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \frac{73}{10}$$
$$b = \frac{219}{2}$$
$$c = -3285$$
, then
D = b^2 - 4 * a * c =
(219/2)^2 - 4 * (73/10) * (-3285) = 431649/4
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 15$$
$$x_{2} = -30$$
Sum and product of roots
[src]
sum
-30 + 15
$$-30 + 15$$
=
-15
$$-15$$
product
-30*15
$$- 450$$
=
-450
$$-450$$
-450