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How to use it?
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Equation 4-6(x+2)=3-5x
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Equation 1+sin(x)=0
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Equation x^2-3*x+1=0
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Identical expressions
- (two *x- six)/(x- three)-(x^ two - six *x+ three)/(x- three)^ two = zero
- (2 multiply by x minus 6) divide by (x minus 3) minus (x squared minus 6 multiply by x plus 3) divide by (x minus 3) squared equally 0
- (two multiply by x minus six) divide by (x minus three) minus (x to the power of two minus six multiply by x plus three) divide by (x minus three) to the power of two equally zero
- (2*x-6)/(x-3)-(x2-6*x+3)/(x-3)2=0
- 2*x-6/x-3-x2-6*x+3/x-32=0
- (2*x-6)/(x-3)-(x²-6*x+3)/(x-3)²=0
- (2*x-6)/(x-3)-(x to the power of 2-6*x+3)/(x-3) to the power of 2=0
- (2x-6)/(x-3)-(x^2-6x+3)/(x-3)^2=0
- (2x-6)/(x-3)-(x2-6x+3)/(x-3)2=0
- 2x-6/x-3-x2-6x+3/x-32=0
- 2x-6/x-3-x^2-6x+3/x-3^2=0
- (2*x-6)/(x-3)-(x^2-6*x+3)/(x-3)^2=O
- (2*x-6) divide by (x-3)-(x^2-6*x+3) divide by (x-3)^2=0
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Similar expressions
- (2*x-6)/(x-3)-(x^2-6*x-3)/(x-3)^2=0
- (2*x-6)/(x-3)-(x^2-6*x+3)/(x+3)^2=0
- (2*x+6)/(x-3)-(x^2-6*x+3)/(x-3)^2=0
- (2*x-6)/(x+3)-(x^2-6*x+3)/(x-3)^2=0
- (2*x-6)/(x-3)+(x^2-6*x+3)/(x-3)^2=0
- (2*x-6)/(x-3)-(x^2+6*x+3)/(x-3)^2=0
(2*x-6)/(x-3)-(x^2-6*x+3)/(x-3)^2=0 equation
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The solution
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[src]
2
2*x - 6 x - 6*x + 3
------- - ------------ = 0
x - 3 2
(x - 3)
$$- \frac{\left(x^{2} - 6 x\right) + 3}{\left(x - 3\right)^{2}} + \frac{2 x - 6}{x - 3} = 0$$
Detail solution
Given the equation:
$$- \frac{\left(x^{2} - 6 x\right) + 3}{\left(x - 3\right)^{2}} + \frac{2 x - 6}{x - 3} = 0$$
Multiply the equation sides by the denominators:
(-3 + x)^2
we get:
$$\left(x - 3\right)^{2} \left(- \frac{\left(x^{2} - 6 x\right) + 3}{\left(x - 3\right)^{2}} + \frac{2 x - 6}{x - 3}\right) = 0$$
$$x^{2} - 6 x + 15 = 0$$
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
$$a = 1$$
$$b = -6$$
$$c = 15$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$x_{1} = 3 + \sqrt{6} i$$
$$x_{2} = 3 - \sqrt{6} i$$
$$- \frac{\left(x^{2} - 6 x\right) + 3}{\left(x - 3\right)^{2}} + \frac{2 x - 6}{x - 3} = 0$$
Multiply the equation sides by the denominators:
(-3 + x)^2
we get:
$$\left(x - 3\right)^{2} \left(- \frac{\left(x^{2} - 6 x\right) + 3}{\left(x - 3\right)^{2}} + \frac{2 x - 6}{x - 3}\right) = 0$$
$$x^{2} - 6 x + 15 = 0$$
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
$$a = 1$$
$$b = -6$$
$$c = 15$$
, then
D = b^2 - 4 * a * c =
(-6)^2 - 4 * (1) * (15) = -24
Because D<0, then the equation
has no real roots,
but complex roots is exists.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 3 + \sqrt{6} i$$
$$x_{2} = 3 - \sqrt{6} i$$
Rapid solution
[src]
___ x1 = 3 - I*\/ 6
$$x_{1} = 3 - \sqrt{6} i$$
___ x2 = 3 + I*\/ 6
$$x_{2} = 3 + \sqrt{6} i$$
x2 = 3 + sqrt(6)*i
Sum and product of roots
[src]
sum
___ ___ 3 - I*\/ 6 + 3 + I*\/ 6
$$\left(3 - \sqrt{6} i\right) + \left(3 + \sqrt{6} i\right)$$
=
6
$$6$$
product
/ ___\ / ___\ \3 - I*\/ 6 /*\3 + I*\/ 6 /
$$\left(3 - \sqrt{6} i\right) \left(3 + \sqrt{6} i\right)$$
=
15
$$15$$
15
Numerical answer
[src]
x1 = 3.0 - 2.44948974278318*i
x2 = 3.0 + 2.44948974278318*i
x2 = 3.0 + 2.44948974278318*i