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(x-2)(x-3)(x-5)-(x-2)(x-4)(x-5)=0 equation
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The solution
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[src]
(x - 2)*(x - 3)*(x - 5) - (x - 2)*(x - 4)*(x - 5) = 0
$$- \left(x - 4\right) \left(x - 2\right) \left(x - 5\right) + \left(x - 3\right) \left(x - 2\right) \left(x - 5\right) = 0$$
Detail solution
Expand the expression in the equation
$$- \left(x - 4\right) \left(x - 2\right) \left(x - 5\right) + \left(x - 3\right) \left(x - 2\right) \left(x - 5\right) = 0$$
We get the quadratic equation
$$x^{2} - 7 x + 10 = 0$$
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
$$a = 1$$
$$b = -7$$
$$c = 10$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 5$$
$$x_{2} = 2$$
$$- \left(x - 4\right) \left(x - 2\right) \left(x - 5\right) + \left(x - 3\right) \left(x - 2\right) \left(x - 5\right) = 0$$
We get the quadratic equation
$$x^{2} - 7 x + 10 = 0$$
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
$$a = 1$$
$$b = -7$$
$$c = 10$$
, then
D = b^2 - 4 * a * c =
(-7)^2 - 4 * (1) * (10) = 9
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 5$$
$$x_{2} = 2$$
Sum and product of roots
[src]
sum
2 + 5
$$2 + 5$$
=
7
$$7$$
product
2*5
$$2 \cdot 5$$
=
10
$$10$$
10