Mister Exam
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x+3
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Similar expressions
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(sqrt(-1-4*x-x*x))=x+3 equation
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The solution
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[src]
________________ \/ -1 - 4*x - x*x = x + 3
$$\sqrt{- x x + \left(- 4 x - 1\right)} = x + 3$$
Detail solution
Given the equation
$$\sqrt{- x x + \left(- 4 x - 1\right)} = x + 3$$
$$\sqrt{- x^{2} - 4 x - 1} = x + 3$$
We raise the equation sides to 2-th degree
$$- x^{2} - 4 x - 1 = \left(x + 3\right)^{2}$$
$$- x^{2} - 4 x - 1 = x^{2} + 6 x + 9$$
Transfer the right side of the equation left part with negative sign
$$- 2 x^{2} - 10 x - 10 = 0$$
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
$$a = -2$$
$$b = -10$$
$$c = -10$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{5}{2} - \frac{\sqrt{5}}{2}$$
$$x_{2} = - \frac{5}{2} + \frac{\sqrt{5}}{2}$$
Because
$$\sqrt{- x^{2} - 4 x - 1} = x + 3$$
and
$$\sqrt{- x^{2} - 4 x - 1} \geq 0$$
then
$$x + 3 \geq 0$$
or
$$-3 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = - \frac{5}{2} + \frac{\sqrt{5}}{2}$$
$$\sqrt{- x x + \left(- 4 x - 1\right)} = x + 3$$
$$\sqrt{- x^{2} - 4 x - 1} = x + 3$$
We raise the equation sides to 2-th degree
$$- x^{2} - 4 x - 1 = \left(x + 3\right)^{2}$$
$$- x^{2} - 4 x - 1 = x^{2} + 6 x + 9$$
Transfer the right side of the equation left part with negative sign
$$- 2 x^{2} - 10 x - 10 = 0$$
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
$$a = -2$$
$$b = -10$$
$$c = -10$$
, then
D = b^2 - 4 * a * c =
(-10)^2 - 4 * (-2) * (-10) = 20
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{5}{2} - \frac{\sqrt{5}}{2}$$
$$x_{2} = - \frac{5}{2} + \frac{\sqrt{5}}{2}$$
Because
$$\sqrt{- x^{2} - 4 x - 1} = x + 3$$
and
$$\sqrt{- x^{2} - 4 x - 1} \geq 0$$
then
$$x + 3 \geq 0$$
or
$$-3 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = - \frac{5}{2} + \frac{\sqrt{5}}{2}$$
Sum and product of roots
[src]
sum
___ 5 \/ 5 - - + ----- 2 2
$$- \frac{5}{2} + \frac{\sqrt{5}}{2}$$
=
___ 5 \/ 5 - - + ----- 2 2
$$- \frac{5}{2} + \frac{\sqrt{5}}{2}$$
product
___ 5 \/ 5 - - + ----- 2 2
$$- \frac{5}{2} + \frac{\sqrt{5}}{2}$$
=
___ 5 \/ 5 - - + ----- 2 2
$$- \frac{5}{2} + \frac{\sqrt{5}}{2}$$
-5/2 + sqrt(5)/2
Rapid solution
[src]
___
5 \/ 5
x1 = - - + -----
2 2
$$x_{1} = - \frac{5}{2} + \frac{\sqrt{5}}{2}$$
x1 = -5/2 + sqrt(5)/2