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Similar expressions
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(2x^2+7x+3)/(x^2-9)=1 equation
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The solution
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2
2*x + 7*x + 3
-------------- = 1
2
x - 9
$$\frac{\left(2 x^{2} + 7 x\right) + 3}{x^{2} - 9} = 1$$
Detail solution
Given the equation:
$$\frac{\left(2 x^{2} + 7 x\right) + 3}{x^{2} - 9} = 1$$
Multiply the equation sides by the denominators:
-9 + x^2
we get:
$$\frac{\left(x^{2} - 9\right) \left(\left(2 x^{2} + 7 x\right) + 3\right)}{x^{2} - 9} = x^{2} - 9$$
$$2 x^{2} + 7 x + 3 = x^{2} - 9$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$2 x^{2} + 7 x + 3 = x^{2} - 9$$
to
$$x^{2} + 7 x + 12 = 0$$
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
$$a = 1$$
$$b = 7$$
$$c = 12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = -3$$
$$x_{2} = -4$$
$$\frac{\left(2 x^{2} + 7 x\right) + 3}{x^{2} - 9} = 1$$
Multiply the equation sides by the denominators:
-9 + x^2
we get:
$$\frac{\left(x^{2} - 9\right) \left(\left(2 x^{2} + 7 x\right) + 3\right)}{x^{2} - 9} = x^{2} - 9$$
$$2 x^{2} + 7 x + 3 = x^{2} - 9$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$2 x^{2} + 7 x + 3 = x^{2} - 9$$
to
$$x^{2} + 7 x + 12 = 0$$
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
$$a = 1$$
$$b = 7$$
$$c = 12$$
, then
D = b^2 - 4 * a * c =
(7)^2 - 4 * (1) * (12) = 1
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -3$$
$$x_{2} = -4$$