Mister Exam
Other calculators
-
How to use it?
- Equation:
- Equation x-y=7
-
Equation 1/(x-1)^2+2/(x-1)-3=0
-
Equation 1/(x+6)=2
-
Equation (x-4)^2-x^2=0
- Express {x} in terms of y where:
- 7*x-1*y=-7
- 10*x-15*y=16
- -14*x-12*y=5
- 13*x-12*y=19
- Derivative of:
-
x^4-2*x^2
- Graphing y =:
-
x^4-2*x^2
- Limit of the function:
-
x^4-2*x^2
-
Identical expressions
- x^ four - two *x^ two =y^ two + two *(|y|)
- x to the power of 4 minus 2 multiply by x squared equally y squared plus 2 multiply by ( module of y|)
- x to the power of four minus two multiply by x to the power of two equally y to the power of two plus two multiply by ( module of y|)
- x4-2*x2=y2+2*(|y|)
- x4-2*x2=y2+2*|y|
- x⁴-2*x²=y²+2*(|y|)
- x to the power of 4-2*x to the power of 2=y to the power of 2+2*(|y|)
- x^4-2x^2=y^2+2(|y|)
- x4-2x2=y2+2(|y|)
- x4-2x2=y2+2|y|
- x^4-2x^2=y^2+2|y|
-
Similar expressions
- x^4-2*x^2=y^2-2*(|y|)
- x^4+2*x^2=y^2+2*(|y|)
x^4-2*x^2=y^2+2*(|y|) equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} - 2 x^{2} = y^{2} + 2 \left|{y}\right|$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - y^{2} - 2 v - 2 \left|{y}\right| = 0$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$x^{4} - 2 x^{2} = y^{2} + 2 \left|{y}\right|$$
to
$$v^{2} - y^{2} - 2 v - 2 \left|{y}\right| = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = - y^{2} - 2 \left|{y}\right|$$
, then
$$D = b^2 - 4\ a\ c = $$
$$- 1 \cdot 4 \left(- y^{2} - 2 \left|{y}\right|\right) + \left(-2\right)^{2} = 4 y^{2} + 8 \left|{y}\right| + 4$$
The equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1$$
Simplify
$$v_{2} = - \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1$$
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:
$$x_{1} = \frac{1 \left(\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = \sqrt{\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
$$x_{2} = \frac{\left(-1\right) \left(\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
$$x_{3} = \frac{1 \left(- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = \sqrt{- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
$$x_{4} = \frac{\left(-1\right) \left(- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
$$x^{4} - 2 x^{2} = y^{2} + 2 \left|{y}\right|$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - y^{2} - 2 v - 2 \left|{y}\right| = 0$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$x^{4} - 2 x^{2} = y^{2} + 2 \left|{y}\right|$$
to
$$v^{2} - y^{2} - 2 v - 2 \left|{y}\right| = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = - y^{2} - 2 \left|{y}\right|$$
, then
$$D = b^2 - 4\ a\ c = $$
$$- 1 \cdot 4 \left(- y^{2} - 2 \left|{y}\right|\right) + \left(-2\right)^{2} = 4 y^{2} + 8 \left|{y}\right| + 4$$
The equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1$$
Simplify
$$v_{2} = - \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1$$
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:
$$x_{1} = \frac{1 \left(\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = \sqrt{\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
$$x_{2} = \frac{\left(-1\right) \left(\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{\frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
$$x_{3} = \frac{1 \left(- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = \sqrt{- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
$$x_{4} = \frac{\left(-1\right) \left(- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{- \frac{\sqrt{4 y^{2} + 8 \left|{y}\right| + 4}}{2} + 1}$$
The graph
Sum and product of roots
[src]
sum
_________________________ _________________________ _________________________ _________________________
/ ________________ / ________________ / ________________ / ________________
/ / 2 / / 2 / / 2 / / 2
-\/ 1 - \/ 1 + y + 2*|y| + \/ 1 - \/ 1 + y + 2*|y| + -\/ 1 + \/ 1 + y + 2*|y| + \/ 1 + \/ 1 + y + 2*|y|
$$\left(- \sqrt{- \sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right) + \left(\sqrt{- \sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right) + \left(- \sqrt{\sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right) + \left(\sqrt{\sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right)$$
=
0
$$0$$
product
_________________________ _________________________ _________________________ _________________________
/ ________________ / ________________ / ________________ / ________________
/ / 2 / / 2 / / 2 / / 2
-\/ 1 - \/ 1 + y + 2*|y| * \/ 1 - \/ 1 + y + 2*|y| * -\/ 1 + \/ 1 + y + 2*|y| * \/ 1 + \/ 1 + y + 2*|y|
$$\left(- \sqrt{- \sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right) * \left(\sqrt{- \sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right) * \left(- \sqrt{\sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right) * \left(\sqrt{\sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}\right)$$
=
2 - y - 2*|y|
$$- y^{2} - 2 \left|{y}\right|$$
Rapid solution
[src]
_________________________
/ ________________
/ / 2
x_1 = -\/ 1 - \/ 1 + y + 2*|y|
$$x_{1} = - \sqrt{- \sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}$$
_________________________
/ ________________
/ / 2
x_2 = \/ 1 - \/ 1 + y + 2*|y|
$$x_{2} = \sqrt{- \sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}$$
_________________________
/ ________________
/ / 2
x_3 = -\/ 1 + \/ 1 + y + 2*|y|
$$x_{3} = - \sqrt{\sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}$$
_________________________
/ ________________
/ / 2
x_4 = \/ 1 + \/ 1 + y + 2*|y|
$$x_{4} = \sqrt{\sqrt{y^{2} + 2 \left|{y}\right| + 1} + 1}$$