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Equation x^4-7x^2+12=0
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Identical expressions
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Similar expressions
- x^4+7x^2+12=0
- x^4-7x^2-12=0
x^4-7x^2+12=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} - 7 x^{2} + 12 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 7 v + 12 = 0$$
This equation is of the form
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 - it is the discriminant.
Because
$$a = 1$$
$$b = -7$$
$$c = 12$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 4$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 4^{\frac{1}{2}}}{1} = 2$$
$$x_{2} = \frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
$$x_{3} = \frac{0}{1} + \frac{1 \cdot 3^{\frac{1}{2}}}{1} = \sqrt{3}$$
$$x_{4} = \frac{\left(-1\right) 3^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{3}$$
$$x^{4} - 7 x^{2} + 12 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 7 v + 12 = 0$$
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 - it is the discriminant.
Because
$$a = 1$$
$$b = -7$$
$$c = 12$$
, then
D = b^2 - 4 * a * c =
(-7)^2 - 4 * (1) * (12) = 1
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 4$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 4^{\frac{1}{2}}}{1} = 2$$
$$x_{2} = \frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
$$x_{3} = \frac{0}{1} + \frac{1 \cdot 3^{\frac{1}{2}}}{1} = \sqrt{3}$$
$$x_{4} = \frac{\left(-1\right) 3^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{3}$$
Sum and product of roots
[src]
sum
___ ___ 0 - 2 + 2 - \/ 3 + \/ 3
$$\left(- \sqrt{3} + \left(\left(-2 + 0\right) + 2\right)\right) + \sqrt{3}$$
=
0
$$0$$
product
___ ___ 1*-2*2*-\/ 3 *\/ 3
$$\sqrt{3} - \sqrt{3} 1 \left(-2\right) 2$$
=
12
$$12$$
12
Rapid solution
[src]
x1 = -2
$$x_{1} = -2$$
x2 = 2
$$x_{2} = 2$$
___ x3 = -\/ 3
$$x_{3} = - \sqrt{3}$$
___ x4 = \/ 3
$$x_{4} = \sqrt{3}$$
The graph