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Identical expressions
- five , nine (x− ten)(x+ thirty-three)= zero .
- 5,9(x−10)(x plus 33) equally 0.
- five , nine (x− ten)(x plus thirty minus three) equally zero .
- 5,9x−10x+33=0.
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-
Similar expressions
- 5,9(x−10)(x-33)=0.
5,9(x−10)(x+33)=0. equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
59*(x - 10)
-----------*(x + 33) = 0
10
$$\frac{59 \left(x - 10\right)}{10} \left(x + 33\right) = 0$$
Detail solution
Expand the expression in the equation
$$\frac{59 \left(x - 10\right)}{10} \left(x + 33\right) = 0$$
We get the quadratic equation
$$\frac{59 x^{2}}{10} + \frac{1357 x}{10} - 1947 = 0$$
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
$$a = \frac{59}{10}$$
$$b = \frac{1357}{10}$$
$$c = -1947$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 10$$
$$x_{2} = -33$$
$$\frac{59 \left(x - 10\right)}{10} \left(x + 33\right) = 0$$
We get the quadratic equation
$$\frac{59 x^{2}}{10} + \frac{1357 x}{10} - 1947 = 0$$
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
$$a = \frac{59}{10}$$
$$b = \frac{1357}{10}$$
$$c = -1947$$
, then
D = b^2 - 4 * a * c =
(1357/10)^2 - 4 * (59/10) * (-1947) = 6436369/100
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 10$$
$$x_{2} = -33$$
Sum and product of roots
[src]
sum
-33 + 10
$$-33 + 10$$
=
-23
$$-23$$
product
-33*10
$$- 330$$
=
-330
$$-330$$
-330