Mister Exam
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How to use it?
- Equation:
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Equation x^5-x^4-2x^3+2x^2-3x+3=0
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- Graphing y =:
- 2x^4-3x^2
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Identical expressions
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Similar expressions
- 2x^4+3x^2=0
2x^4-3x^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$2 x^{4} - 3 x^{2} = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$2 v^{2} - 3 v = 0$$
This equation is of the form
$$a\ v^2 + b\ v + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 2$$
$$b = -3$$
$$c = 0$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 2 \cdot 4 \cdot 0 + \left(-3\right)^{2} = 9$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = \frac{3}{2}$$
Simplify
$$v_{2} = 0$$
Simplify
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = \frac{0}{1} + \frac{1 \left(\frac{3}{2}\right)^{\frac{1}{2}}}{1} = \frac{\sqrt{6}}{2}$$
$$x_{2} = \frac{\left(-1\right) \left(\frac{3}{2}\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = - \frac{\sqrt{6}}{2}$$
$$x_{3} = \frac{1 \cdot 0^{\frac{1}{2}}}{1} + \frac{0}{1} = 0$$
$$2 x^{4} - 3 x^{2} = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$2 v^{2} - 3 v = 0$$
This equation is of the form
$$a\ v^2 + b\ v + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 2$$
$$b = -3$$
$$c = 0$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 2 \cdot 4 \cdot 0 + \left(-3\right)^{2} = 9$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = \frac{3}{2}$$
Simplify
$$v_{2} = 0$$
Simplify
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = \frac{0}{1} + \frac{1 \left(\frac{3}{2}\right)^{\frac{1}{2}}}{1} = \frac{\sqrt{6}}{2}$$
$$x_{2} = \frac{\left(-1\right) \left(\frac{3}{2}\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = - \frac{\sqrt{6}}{2}$$
$$x_{3} = \frac{1 \cdot 0^{\frac{1}{2}}}{1} + \frac{0}{1} = 0$$
Sum and product of roots
[src]
sum
___ ___
-\/ 6 \/ 6
0 + ------- + -----
2 2
$$\left(0\right) + \left(- \frac{\sqrt{6}}{2}\right) + \left(\frac{\sqrt{6}}{2}\right)$$
=
0
$$0$$
product
___ ___
-\/ 6 \/ 6
0 * ------- * -----
2 2
$$\left(0\right) * \left(- \frac{\sqrt{6}}{2}\right) * \left(\frac{\sqrt{6}}{2}\right)$$
=
0
$$0$$
Rapid solution
[src]
x_1 = 0
$$x_{1} = 0$$
___
-\/ 6
x_2 = -------
2
$$x_{2} = - \frac{\sqrt{6}}{2}$$
___
\/ 6
x_3 = -----
2
$$x_{3} = \frac{\sqrt{6}}{2}$$
The graph