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sqrt(4+2x−x^2)=x−2. equation
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The solution
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______________ / 2 \/ 4 + 2*x - x = x - 2
$$\sqrt{- x^{2} + 2 x + 4} = x - 2$$
Detail solution
Given the equation
$$\sqrt{- x^{2} + 2 x + 4} = x - 2$$
$$\sqrt{- x^{2} + 2 x + 4} = x - 2$$
We raise the equation sides to 2-th degree
$$- x^{2} + 2 x + 4 = \left(x - 2\right)^{2}$$
$$- x^{2} + 2 x + 4 = x^{2} - 4 x + 4$$
Transfer the right side of the equation left part with negative sign
$$- 2 x^{2} + 6 x = 0$$
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
$$a = -2$$
$$b = 6$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 0$$
Simplify
$$x_{2} = 3$$
Simplify
Because
$$\sqrt{- x^{2} + 2 x + 4} = x - 2$$
and
$$\sqrt{- x^{2} + 2 x + 4} \geq 0$$
then
$$x - 2 \geq 0$$
or
$$2 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 3$$
$$\sqrt{- x^{2} + 2 x + 4} = x - 2$$
$$\sqrt{- x^{2} + 2 x + 4} = x - 2$$
We raise the equation sides to 2-th degree
$$- x^{2} + 2 x + 4 = \left(x - 2\right)^{2}$$
$$- x^{2} + 2 x + 4 = x^{2} - 4 x + 4$$
Transfer the right side of the equation left part with negative sign
$$- 2 x^{2} + 6 x = 0$$
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
$$a = -2$$
$$b = 6$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(6)^2 - 4 * (-2) * (0) = 36
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 0$$
Simplify
$$x_{2} = 3$$
Simplify
Because
$$\sqrt{- x^{2} + 2 x + 4} = x - 2$$
and
$$\sqrt{- x^{2} + 2 x + 4} \geq 0$$
then
$$x - 2 \geq 0$$
or
$$2 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 3$$
The graph