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Identical expressions
- ((x^ two + four x- four)/(x^ two - twenty-five))/((2x+4)/(6x+ thirty))= zero
- ((x squared plus 4x minus 4) divide by (x squared minus 25)) divide by ((2x plus 4) divide by (6x plus 30)) equally 0
- ((x to the power of two plus four x minus four) divide by (x to the power of two minus twenty minus five)) divide by ((2x plus 4) divide by (6x plus thirty)) equally zero
- ((x2+4x-4)/(x2-25))/((2x+4)/(6x+30))=0
- x2+4x-4/x2-25/2x+4/6x+30=0
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- x^2+4x-4/x^2-25/2x+4/6x+30=0
- ((x^2+4x-4)/(x^2-25))/((2x+4)/(6x+30))=O
- ((x^2+4x-4) divide by (x^2-25)) divide by ((2x+4) divide by (6x+30))=0
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Similar expressions
- ((x^2+4x-4)/(x^2-25))/((2x+4)/(6x-30))=0
- ((x^2-4x-4)/(x^2-25))/((2x+4)/(6x+30))=0
- ((x^2+4x+4)/(x^2-25))/((2x+4)/(6x+30))=0
- ((x^2+4x-4)/(x^2-25))/((2x-4)/(6x+30))=0
- ((x^2+4x-4)/(x^2+25))/((2x+4)/(6x+30))=0
((x^2+4x-4)/(x^2-25))/((2x+4)/(6x+30))=0 equation
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The solution
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[src]
/ 2 \ |x + 4*x - 4| |------------| | 2 | \ x - 25 / -------------- = 0 /2*x + 4 \ |--------| \6*x + 30/
$$\frac{\frac{1}{x^{2} - 25} \left(\left(x^{2} + 4 x\right) - 4\right)}{\left(2 x + 4\right) \frac{1}{6 x + 30}} = 0$$
Detail solution
Given the equation:
$$\frac{\frac{1}{x^{2} - 25} \left(\left(x^{2} + 4 x\right) - 4\right)}{\left(2 x + 4\right) \frac{1}{6 x + 30}} = 0$$
transform:
Take common factor from the equation
$$\frac{3 \left(x^{2} + 4 x - 4\right)}{\left(x - 5\right) \left(x + 2\right)} = 0$$
the denominator
$$x - 5$$
then
the denominator
$$x + 2$$
then
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$3 x^{2} + 12 x - 12 = 0$$
solve the resulting equation:
1.
$$3 x^{2} + 12 x - 12 = 0$$
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
$$a = 3$$
$$b = 12$$
$$c = -12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = -2 + 2 \sqrt{2}$$
$$x_{2} = - 2 \sqrt{2} - 2$$
but
The final answer:
$$x_{1} = -2 + 2 \sqrt{2}$$
$$x_{2} = - 2 \sqrt{2} - 2$$
$$\frac{\frac{1}{x^{2} - 25} \left(\left(x^{2} + 4 x\right) - 4\right)}{\left(2 x + 4\right) \frac{1}{6 x + 30}} = 0$$
transform:
Take common factor from the equation
$$\frac{3 \left(x^{2} + 4 x - 4\right)}{\left(x - 5\right) \left(x + 2\right)} = 0$$
the denominator
$$x - 5$$
then
x is not equal to 5
the denominator
$$x + 2$$
then
x is not equal to -2
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$3 x^{2} + 12 x - 12 = 0$$
solve the resulting equation:
1.
$$3 x^{2} + 12 x - 12 = 0$$
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
$$a = 3$$
$$b = 12$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(12)^2 - 4 * (3) * (-12) = 288
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -2 + 2 \sqrt{2}$$
$$x_{2} = - 2 \sqrt{2} - 2$$
but
x is not equal to 5
x is not equal to -2
The final answer:
$$x_{1} = -2 + 2 \sqrt{2}$$
$$x_{2} = - 2 \sqrt{2} - 2$$
Rapid solution
[src]
___ x1 = -2 + 2*\/ 2
$$x_{1} = -2 + 2 \sqrt{2}$$
___ x2 = -2 - 2*\/ 2
$$x_{2} = - 2 \sqrt{2} - 2$$
x2 = -2*sqrt(2) - 2
Sum and product of roots
[src]
sum
___ ___ -2 + 2*\/ 2 + -2 - 2*\/ 2
$$\left(- 2 \sqrt{2} - 2\right) + \left(-2 + 2 \sqrt{2}\right)$$
=
-4
$$-4$$
product
/ ___\ / ___\ \-2 + 2*\/ 2 /*\-2 - 2*\/ 2 /
$$\left(-2 + 2 \sqrt{2}\right) \left(- 2 \sqrt{2} - 2\right)$$
=
-4
$$-4$$
-4