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- x^4-24x^2+25=0
- x^4+24x^2-25=0
x^4-24x^2-25=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} - 24 x^{2} - 25 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 24 v - 25 = 0$$
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
$$a = 1$$
$$b = -24$$
$$c = -25$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 1 \cdot 4 \left(-25\right) + \left(-24\right)^{2} = 676$$
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
$$v_{1} = 25$$
Simplify
$$v_{2} = -1$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 25^{\frac{1}{2}}}{1} = 5$$
$$x_{2} = \frac{\left(-1\right) 25^{\frac{1}{2}}}{1} + \frac{0}{1} = -5$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-1\right)^{\frac{1}{2}}}{1} = i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-1\right)^{\frac{1}{2}}}{1} = - i$$
$$x^{4} - 24 x^{2} - 25 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 24 v - 25 = 0$$
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
$$a = 1$$
$$b = -24$$
$$c = -25$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 1 \cdot 4 \left(-25\right) + \left(-24\right)^{2} = 676$$
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
$$v_{1} = 25$$
Simplify
$$v_{2} = -1$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 25^{\frac{1}{2}}}{1} = 5$$
$$x_{2} = \frac{\left(-1\right) 25^{\frac{1}{2}}}{1} + \frac{0}{1} = -5$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-1\right)^{\frac{1}{2}}}{1} = i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-1\right)^{\frac{1}{2}}}{1} = - i$$
Sum and product of roots
[src]
sum
-5 + 5 + -I + I
$$\left(-5\right) + \left(5\right) + \left(- i\right) + \left(i\right)$$
=
0
$$0$$
product
-5 * 5 * -I * I
$$\left(-5\right) * \left(5\right) * \left(- i\right) * \left(i\right)$$
=
-25
$$-25$$
Rapid solution
[src]
x_1 = -5
$$x_{1} = -5$$
x_2 = 5
$$x_{2} = 5$$
x_3 = -I
$$x_{3} = - i$$
x_4 = I
$$x_{4} = i$$