Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 81*x^3+18*x^2+x=0
-
Equation 6=8-5(7x-1)
-
Equation 3x=28-x
-
Equation 3x=2
- Express {x} in terms of y where:
- -13*x-7*y=16
- -16*x+10*y=-11
- -2*x-15*y=-1
- -14*x+5*y=-1
-
Identical expressions
- (y^ four + four *y^ two - twelve)/ four = zero
- (y to the power of 4 plus 4 multiply by y squared minus 12) divide by 4 equally 0
- (y to the power of four plus four multiply by y to the power of two minus twelve) divide by four equally zero
- (y4+4*y2-12)/4=0
- y4+4*y2-12/4=0
- (y⁴+4*y²-12)/4=0
- (y to the power of 4+4*y to the power of 2-12)/4=0
- (y^4+4y^2-12)/4=0
- (y4+4y2-12)/4=0
- y4+4y2-12/4=0
- y^4+4y^2-12/4=0
- (y^4+4*y^2-12)/4=O
- (y^4+4*y^2-12) divide by 4=0
-
Similar expressions
- (y^4-4*y^2-12)/4=0
- (y^4+4*y^2+12)/4=0
(y^4+4*y^2-12)/4=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
4 2
y + 4*y - 12
-------------- = 0
4
$$\frac{\left(y^{4} + 4 y^{2}\right) - 12}{4} = 0$$
Detail solution
Given the equation:
$$\frac{\left(y^{4} + 4 y^{2}\right) - 12}{4} = 0$$
Do replacement
$$v = y^{2}$$
then the equation will be the:
$$\frac{v^{2}}{4} + v - 3 = 0$$
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
$$a = \frac{1}{4}$$
$$b = 1$$
$$c = -3$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 2$$
$$v_{2} = -6$$
The final answer:
Because
$$v = y^{2}$$
then
$$y_{1} = \sqrt{v_{1}}$$
$$y_{2} = - \sqrt{v_{1}}$$
$$y_{3} = \sqrt{v_{2}}$$
$$y_{4} = - \sqrt{v_{2}}$$
then:
$$y_{1} = $$
$$\frac{0}{1} + \frac{2^{\frac{1}{2}}}{1} = \sqrt{2}$$
$$y_{2} = $$
$$\frac{\left(-1\right) 2^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{2}$$
$$y_{3} = $$
$$\frac{0}{1} + \frac{\left(-6\right)^{\frac{1}{2}}}{1} = \sqrt{6} i$$
$$y_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-6\right)^{\frac{1}{2}}}{1} = - \sqrt{6} i$$
$$\frac{\left(y^{4} + 4 y^{2}\right) - 12}{4} = 0$$
Do replacement
$$v = y^{2}$$
then the equation will be the:
$$\frac{v^{2}}{4} + v - 3 = 0$$
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 - it is the discriminant.
Because
$$a = \frac{1}{4}$$
$$b = 1$$
$$c = -3$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1/4) * (-3) = 4
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 2$$
$$v_{2} = -6$$
The final answer:
Because
$$v = y^{2}$$
then
$$y_{1} = \sqrt{v_{1}}$$
$$y_{2} = - \sqrt{v_{1}}$$
$$y_{3} = \sqrt{v_{2}}$$
$$y_{4} = - \sqrt{v_{2}}$$
then:
$$y_{1} = $$
$$\frac{0}{1} + \frac{2^{\frac{1}{2}}}{1} = \sqrt{2}$$
$$y_{2} = $$
$$\frac{\left(-1\right) 2^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{2}$$
$$y_{3} = $$
$$\frac{0}{1} + \frac{\left(-6\right)^{\frac{1}{2}}}{1} = \sqrt{6} i$$
$$y_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-6\right)^{\frac{1}{2}}}{1} = - \sqrt{6} i$$
Sum and product of roots
[src]
sum
___ ___ ___ ___ - \/ 2 + \/ 2 - I*\/ 6 + I*\/ 6
$$\left(\left(- \sqrt{2} + \sqrt{2}\right) - \sqrt{6} i\right) + \sqrt{6} i$$
=
0
$$0$$
product
___ ___ / ___\ ___ -\/ 2 *\/ 2 *\-I*\/ 6 /*I*\/ 6
$$\sqrt{6} i - \sqrt{2} \sqrt{2} \left(- \sqrt{6} i\right)$$
=
-12
$$-12$$
-12
Rapid solution
[src]
___ y1 = -\/ 2
$$y_{1} = - \sqrt{2}$$
___ y2 = \/ 2
$$y_{2} = \sqrt{2}$$
___ y3 = -I*\/ 6
$$y_{3} = - \sqrt{6} i$$
___ y4 = I*\/ 6
$$y_{4} = \sqrt{6} i$$
y4 = sqrt(6)*i
Numerical answer
[src]
y1 = 1.4142135623731
y2 = -2.44948974278318*i
y3 = 2.44948974278318*i
y4 = -1.4142135623731
y4 = -1.4142135623731
The graph