Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^2-9x+20=0
-
Equation 20-7(5x-4)=6
-
Equation (x-2)^2+(x-8)^2=2x^2
-
Equation x^4+2x^2-3=0
- Express {x} in terms of y where:
- 10*x-1*y=10
- 15*x-17*y=19
- 9*x-8*y=6
- 15*x+15*y=16
- Graphing y =:
-
x^4+2x^2-3
-
Identical expressions
- x^ four + two x^2- three = zero
- x to the power of 4 plus 2x squared minus 3 equally 0
- x to the power of four plus two x squared minus three equally zero
- x4+2x2-3=0
- x⁴+2x²-3=0
- x to the power of 4+2x to the power of 2-3=0
- x^4+2x^2-3=O
-
Similar expressions
- x^4-2x^2-3=0
- x^4+2x^2+3=0
x^4+2x^2-3=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} + 2 x^{2} - 3 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} + 2 v - 3 = 0$$
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
$$a = 1$$
$$b = 2$$
$$c = -3$$
, then
$$D = b^2 - 4 * a * c = $$
$$2^{2} - 1 \cdot 4 \left(-3\right) = 16$$
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
$$v_{1} = 1$$
Simplify
$$v_{2} = -3$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = \frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-3\right)^{\frac{1}{2}}}{1} = \sqrt{3} i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-3\right)^{\frac{1}{2}}}{1} = - \sqrt{3} i$$
$$x^{4} + 2 x^{2} - 3 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} + 2 v - 3 = 0$$
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
$$a = 1$$
$$b = 2$$
$$c = -3$$
, then
$$D = b^2 - 4 * a * c = $$
$$2^{2} - 1 \cdot 4 \left(-3\right) = 16$$
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
$$v_{1} = 1$$
Simplify
$$v_{2} = -3$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = \frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-3\right)^{\frac{1}{2}}}{1} = \sqrt{3} i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-3\right)^{\frac{1}{2}}}{1} = - \sqrt{3} i$$
Rapid solution
[src]
x_1 = -1
$$x_{1} = -1$$
x_2 = 1
$$x_{2} = 1$$
___ x_3 = -I*\/ 3
$$x_{3} = - \sqrt{3} i$$
___ x_4 = I*\/ 3
$$x_{4} = \sqrt{3} i$$
Sum and product of roots
[src]
sum
___ ___ -1 + 1 + -I*\/ 3 + I*\/ 3
$$\left(-1\right) + \left(1\right) + \left(- \sqrt{3} i\right) + \left(\sqrt{3} i\right)$$
=
0
$$0$$
product
___ ___ -1 * 1 * -I*\/ 3 * I*\/ 3
$$\left(-1\right) * \left(1\right) * \left(- \sqrt{3} i\right) * \left(\sqrt{3} i\right)$$
=
-3
$$-3$$
Numerical answer
[src]
x1 = 1.0
x2 = 1.73205080756888*i
x3 = -1.73205080756888*i
x4 = -1.0
x4 = -1.0
The graph