Mister Exam
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How to use it?
- Equation:
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Equation x^2+2*x+10=0
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Identical expressions
- sqrt(3x^ two +4x- six)= two
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- √(3x^2+4x-6)=2
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- sqrt3x^2+4x-6=2
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Similar expressions
- sqrt(3x^2-4x-6)=2
- sqrt(3x^2+4x+6)=2
sqrt(3x^2+4x-6)=2 equation
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The solution
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[src]
________________ / 2 \/ 3*x + 4*x - 6 = 2
$$\sqrt{\left(3 x^{2} + 4 x\right) - 6} = 2$$
Detail solution
Given the equation
$$\sqrt{\left(3 x^{2} + 4 x\right) - 6} = 2$$
$$\sqrt{3 x^{2} + 4 x - 6} = 2$$
We raise the equation sides to 2-th degree
$$3 x^{2} + 4 x - 6 = 4$$
$$3 x^{2} + 4 x - 6 = 4$$
Transfer the right side of the equation left part with negative sign
$$3 x^{2} + 4 x - 10 = 0$$
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
$$a = 3$$
$$b = 4$$
$$c = -10$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{2}{3} + \frac{\sqrt{34}}{3}$$
$$x_{2} = - \frac{\sqrt{34}}{3} - \frac{2}{3}$$
Because
$$\sqrt{3 x^{2} + 4 x - 6} = 2$$
and
$$\sqrt{3 x^{2} + 4 x - 6} \geq 0$$
then
$$2 \geq 0$$
The final answer:
$$x_{1} = - \frac{2}{3} + \frac{\sqrt{34}}{3}$$
$$x_{2} = - \frac{\sqrt{34}}{3} - \frac{2}{3}$$
$$\sqrt{\left(3 x^{2} + 4 x\right) - 6} = 2$$
$$\sqrt{3 x^{2} + 4 x - 6} = 2$$
We raise the equation sides to 2-th degree
$$3 x^{2} + 4 x - 6 = 4$$
$$3 x^{2} + 4 x - 6 = 4$$
Transfer the right side of the equation left part with negative sign
$$3 x^{2} + 4 x - 10 = 0$$
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
$$a = 3$$
$$b = 4$$
$$c = -10$$
, then
D = b^2 - 4 * a * c =
(4)^2 - 4 * (3) * (-10) = 136
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{2}{3} + \frac{\sqrt{34}}{3}$$
$$x_{2} = - \frac{\sqrt{34}}{3} - \frac{2}{3}$$
Because
$$\sqrt{3 x^{2} + 4 x - 6} = 2$$
and
$$\sqrt{3 x^{2} + 4 x - 6} \geq 0$$
then
$$2 \geq 0$$
The final answer:
$$x_{1} = - \frac{2}{3} + \frac{\sqrt{34}}{3}$$
$$x_{2} = - \frac{\sqrt{34}}{3} - \frac{2}{3}$$
Rapid solution
[src]
____
2 \/ 34
x1 = - - + ------
3 3
$$x_{1} = - \frac{2}{3} + \frac{\sqrt{34}}{3}$$
____
2 \/ 34
x2 = - - - ------
3 3
$$x_{2} = - \frac{\sqrt{34}}{3} - \frac{2}{3}$$
x2 = -sqrt(34)/3 - 2/3
Sum and product of roots
[src]
sum
____ ____ 2 \/ 34 2 \/ 34 - - + ------ + - - - ------ 3 3 3 3
$$\left(- \frac{\sqrt{34}}{3} - \frac{2}{3}\right) + \left(- \frac{2}{3} + \frac{\sqrt{34}}{3}\right)$$
=
-4/3
$$- \frac{4}{3}$$
product
/ ____\ / ____\ | 2 \/ 34 | | 2 \/ 34 | |- - + ------|*|- - - ------| \ 3 3 / \ 3 3 /
$$\left(- \frac{2}{3} + \frac{\sqrt{34}}{3}\right) \left(- \frac{\sqrt{34}}{3} - \frac{2}{3}\right)$$
=
-10/3
$$- \frac{10}{3}$$
-10/3